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INDUSTRIAL PHYSICS 



MECHANICS 



BY 

Lf RAYMOND SMITH 

INSTRUCTOR IN INDUSTRIAL PHYSICS WM. D. DICKINSON HIGH SCHOOL, 
JERSEY CITY, N. J. MEMBER AMERICAN SOCIETY OF MECHANICAL ENGINEERS 



First Edition 



McGRAW-HILL BOOK COMPANY, Inc. 
NEW YORK: 370 SEVENTH AVENUE 

LONDON: 6 & 8 BOUVERIE ST., E. C. 4 

1922 



'' / 






Copyright, 1922, by the 
McGraw-Hill Book Company, Inc. 



k\iQ-i':j 



TSE MAPLE PRESS TOKk PA 



©CI.AK77H99 
'HO I 






PREFACE 

The present trend in education has created a demand for a 
series of textbooks in which the material presented js more 
closely connected with the every-day life of the student. This 
volume is the result of an attempt to provide a textbook in 
elementary, practical mechanics, — a textbook suitable for use 
in technical, industrial, vocational and evening schools. 

The chief difficulty encountered in writing an educational 
book is the choice of material. The material presented here 
consists largely of notes used in class by the author. It has 
been tried out and has proven successful. Certain time- 
honored topics, usually included in similar books, have been 
omitted; there has been no sacrifice, however, of fundamental 
principles. The order of presentation has been found satis- 
factory. It may be changed, if desired, without impairing 
the value of the course. An attempt has been made through- 
out to keep the diction simple and understandable. 

Attention is directed to the questions and problems at the 
end of each chapter or division. The questions depend upon 
the subject-matter preceding and are invaluable both for 
study and review. The problems are not difficult and, it is 
hoped, sufficiently generous in number. Experience has 
proven that problems are an excellent medium for clarifying 
misunderstandings on the part of the stuUent. Arithmetic 
is sufficient for most solutions, although a knowledge of 
algebra and trigonometry will be helpful. A special chapter 
dealing with elementary trigonometry has been included. 

The author makes no particular claim to originality. He 
has consulted various standard works freely and acknowledges 
his indebtedness to many of them. Care has been exercised to 
keep the book free from errors. In case errors have crept in, 
the author will be glad to have his attention called to them. 



VI PREFACE 

Criticism of scope and content will also be welcomed. 

Certain chapters have been made possible through the 
hearty cooperation of various firms and manufacturing plants 
in furnishing information, photographs, electrotypes, etc. 
Grateful acknowledgment is made to the following: Brown 
and Sharpe Mfg. Co.; Central Scientific Co.; Cole Motor Car 
Co.; Curtiss Aeroplane and Motor Corp.; Dodge Sales and 
Engineering Co.; Edw. R. Ladew Co.; Fafnir Bearing Co.; 
Ford Motor Co.; Goodell-Pratt Co.; Goodyear Rubber Co.; 
Goulds Mfg. Co.; Hyatt Roller Bearing Co.; Keufifel and 
Esser Co.; L. E. Knott and Co.; Link-Belt Co.; L. S. Starrett 
Co.; Skinner Engine Co.; Timken Roller Bearing Co.; Whitney 
Mfg. Co.; Worthington Pump and Machinery Corp.; Yale 
and Towne Mfg. Co. 

The author wishes to thank his colleagues, T. Gilbert 
McFadden, John H. Finn and Albert E. Dickie, for the 
valuable help the}^ have given in the preparation of the manu- 
script. He also wishes to thank Frank E. Mathewson, Direc- 
tor of the Technical and Industrial Department, Wm. L. 
Dickinson High School, for suggestions and advice incident to 
the publication of the book. 

L. R. Smith. 
Jersey City, N. J., 
Jufw, 1922. 



CONTENTS 

Page 

Preface v 

CHAPTER I 

Introductory 1-6 

Definition and scope of physics — Definition and divisions of 
mechanics — Matter and energy defined — Constitution of matter 
— Kinetic theory of matter — Physical and chemical changes — 
Definition of force and equilibrium — General properties of 
matter — Specific properties of matter. 

CHAPTER II 

Measurement and Measuring Instruments 7-19 

English and metric systems of measurement — Metric tables — 
Common English-metric equivalents — The steel rule — Adjust- 
able and non-adjustable protractors — Ordinary calipers — Thq 
speed indicator — The slide rule — The micrometer caliper — The 
vernier caliper. 

CHAPTER III 

Elementary Trigonometry 20-22 

Trigonometry defined and discussed — Trigonometric functions 
— Illustrative trigonometric solutions. 

CHAPTER IV 

Gravitation and Gravity 23-29 

Universal gravitation — ^Law of universal gravitation — Gravity 
— Weight — Center of gravity — Stable, unstable and neutral 
equilibrium. 

CHAPTER V 

Forces . 30-34 

Effect of forces — Definition of a force — Action and reaction — 
Measurement of a force — Graphical representation of forces — 
Moment of a force. 

vii 



viii CONTENTS 

CHAPTER VI 

Page 

Motion ^^^^ 

Motion defined— Absolute and relative motion— Translatory 
and rotary motion— Accelerated motion — Speed and velocity — 
Momentum— Newton's laws of motion— Graphical representa- 
tion of motion — Velocity of rotation — The radian. 

CHAPTER VII 

Composition of Forces and Velocities 44-47 

Resultant of two or more forces— Parallelogram of forces- 
Experimental verification of the parallelogram law— Composi- 
tion of velocities— Directions for graphical work. 

CHAPTER VIII 

Resolution of Forces and Velocities 48-51 

Graphical resolution— Rectangular components determined 
graphically— Use of squared paper— Rectangular components 
determined trigonometrically. 

CHAPTER IX 

Equilibrium OF Concurrent Forces 52-56 

Laws of concurrent forces in equilibrium— Concurrent forces 
applied at the same point— Triangle of forces— Concurrent 
forces applied at separate points. 

CHAPTER X 

Equilibrium of Parallel Forces 57-61 

Tendencies of parallel forces— Laws of parallel forces— Experi- 
mental verification of the laws of parallel forces— Resultant 
and point of application for a system of parallel forces— The 
couple. 

CHAPTER XI 

Equilibrium of Non-concurrent Forces 62-65 

Conditions for equilibrium of non-concurrent forces — The ladder 
— The wall crane. 



CONTENTS IX 

CHAPTER XII 

Page 

Commercial and Laboratory Structures 66-76 

General discussion of trusses — Types of roof trusses — Types of 
bridge trusses — Laboratory study of a roof truss — ^Laboratory 
study of a stick and tie — ^Laboratory study of a hoisting crane 
— ^Laboratory study of shear legs. 

CHAPTER XIII 

Elasticity 77-86 

Discussion and definition of elasticity — Hooke's law — The 
elastic limit — The yield point — Stress — Strain — Young's 
modulus — Ultimate strength — Factor of safety — Elastic fatigue 
— Shear explained — Shearing stress — Shearing strain — Modulus 
of rigidity — Practical illustrations of shear. 

CHAPTER XIV 

Work 87-91 

Definition and explanation of work — Measurement of work — 
Units of work — Time and work — Work diagrams. 

CHAPTER XV 

Power 92-97 

Power defined and discussed — Units of power — Brake horse- 
power — S.A.E. horsepower — Indicated horsepower — Mechan- 
ical efficiency. 

CHAPTER XVI 

Energy 98-102 

Nature of energy — Fixed energy — Kinetic energy — Why a body 
possesses energy — Transformation of energy — Conservation of 
energy — The mechanical equivalent of heat — The British 
thermal unit. 

CHAPTER XVII 

Friction 103-111 

Cause of friction — Advantages and disadvantages of friction 
— Coefficient of friction — ^Laws of friction — Plain bearings — 
Ball bearings — Roller bearings — Lubrication of bearings. 



X CONTENTS 

CHAPTER XVIII 

Page 

Simple Machines 112-122 

Definition and discussion of a machine — Classification of 
machines — Input — Output — Efficiency — Velocity ratio — Mech- 
anical advantage — ^Law of frictionless machines — Law of 
non-frictionless machines — First, second and third class levers — 
Pulleys — The wheel and axle — The inclined plane — The screw — 
The wedge. 

CHAPTER XIX 

Practical Study of Machines 123-139 

Types of chain blocks — ^Lalx)ratory study of a differential chain 
block — Laboratory study of a screw-geared chain block — 
The spur-geared chain block — Laboratory study of a chain-drive 
bicycle — Laboratory study of a jack screw — The transmission 
of an automobile. 

CHAPTER XX 

Mechanical Transmission of Power 140-160 

Methods of power transmission — Shafts — Couplings — Universal 
couplings — Clutches — Cams — Links — Screw threads — Chains 
and sprockets — Pulleys and belts — Pulleys and ropes — Friction 
wheels — Toothed gears — Spur gears — Bevel gears — Worm 
gears — Helical gears — Spiral gears. 

CHAPTER XXI 

Fluids 161-201 

The three forms of matter — Density — Specific gravity — Pres- 
sure defined — The laws of liquid pressure — Transmission of 
pressure by liquids — Pascal's law stated and illustrated — The 
hydraulic press — Communicating columns — Pressure and weight 
distinguished — Total liquid pressure on plane surfaces — Center 
of pressure — Dams and retaining walls — Waterwheels — 
Buoyant force of liquids — Archimedes' principle — The hydro- 
meter — Characteristics of gases — The atmosphere — Proof that 
air has weight — Proof that air exerts pressure — Torricelli's 
experiment — Pascal's experiment — The mercury barometer — 
The aneroid barometer — The barograph — Standard conditions 
of pressure and temperature — The Magdeburg hemispheres — 



CONTENTS XI 

Page 
Boyle's law stated and illustrated — The aeroplane — The balloon 
— The siphon — The air pump — The air brake — Pumps for 
liquids — The pressure gauge — The vacuum gauge — 
Manometers. 

CHAPTER XXII 

Falling Bodies; Centrifugal Force; the Pendulum . . . 202-214 
The effect of gravity on falling bodies — The acceleration of 
gravity — Freely falling bodies — Bodies rolling down an incline 
— Bodies projected vertically downward — Bodies projected ver- 
tically upward — Bodies projected horizontally — Bodies pro- 
jected at an angle of elevation of less than 90° — The nature of 
centrifugal force — Effects and uses of centrifugal force — The 
formula for centrifugal force — The pendulum defined and dis- 
cussed — Why a pendulum vibrates — ^Laws of the pendulum. 

Appendix 215-218 

Useful information — English and metric equivalents — 
Density of common substances — Units frequently used — 
Specific gravity of common substances — Tensile strength, 
compressive strength and shearing strength for cast iron, 
wrought iron and mild steel — Modulus of elasticity (tension, 
compression and shear) for cast iron, wrought iron and mild 
steel — Trigonometric tables — Decimal equivalents of parts 
of an inch. 



INDUSTRIAL PHYSICS 



MECHANICS 

CHAPTER I 
INTRODUCTORY 

1. Physics. — Physics is often defined as the science of 
7natter and energy ^ — the science which attempts to explain the 
relation between physical phenomena and the causes producing 
them. Physics is conveniently divided into five parts: 
(1) mechanics; (2) heat; (3) light; (4) sound; (5) electricity. 
This volume deals with mechanics only, a knowledge of which 
is very useful in studying the remaining divisions of the 
subject. 

2. Industrial Physics. — Industrial physics differs from the 
so-called academic physics in that its subject matter is more 
practical. It substitutes material of live value for certain 
time-honored topics which seem to have little or no connection 
with the life of the pupil. It is intended for serious, energetic 
students who wish to get a better training than has been 
possible with the material customarily presented. 

3. Mechanics. — Mechanics is that branch of physical science 
which deals with forces and their effects. It includes a study 
of the action of forces on solids, liquids and gases. Mechanics 
may be divided into two distinct parts : statics and kinetics. 

Statics deals with bodies so acted upon by forces that no 
motion results. 

Kinetics deals with bodies so acted upon by forces that 
motion results. 

1 



2 MECHANICS 

4. Matter. — Matter is anything which has weight or which 
takes up room. Matter may exist in three different forms: 
solids J liquids or gases. Any material substance, such as steel, 
gold, water, air, etc., is composed of matter. 

5. Energy. — Energy is the ability to do work. Gasoline 
possesses energ}^ because it enables an internal combustion 
engine to do work. 

6. The Molecule. — All matter is thought to consist of 
small particles called molecules. A molecule is defined as 
the smallest particle into which a substance may be divided 
without destroying its identity. For example, a molecule of 
water is written H2O; two molecules of water is written 
2H2O, etc. Suppose we have one hundred molecules of 
water or 100 HoO; we can then divide it into 100 parts, each 
part being H2O and having the characteristics of water. If 
we go beyond this in our division, we break up the molecule 
into oxygen and hydrogen, two gases which in no way resemble 
the original water. 

Molecules are so minute that they are not discernible with 
the most powerful microscopes. It has been estimated by 
Lord Kelvin^ that, if a quantity of water the size of a foot- 
ball were magnified to the size of the earth, the molecules 
would be between small shot and footballs in size. 

7. The Atom. — A molecule is made up of atoms. By 
breaking up the water molecule (HoO) we get the atoms 
H, H and 0. These atoms in no way resemble the molecule 
of which they were formerly a part. The atom then, after 
the molecule, is the next step in the division of matter and is 
the smallest particle of matter capable of entering into combination. 

8. The Electron. — According to the most recent investiga- 
tions, the atom is a very complex structure. It is thought to 
be composed of a nucleus or positively charged vibrating 
particle surrounded by various negatively charged particles. 

iSir William Thomson (1824-1907). One of the greatest of the 
nineteenth-century physicists. Born in Ireland. P^ifty-three years 
professor of physics at Glasgow University. Noted for his work in heat 
and electricity. 



INTRODUCTORY 3 

These negative particles, called electrons^ are very mobile and 
possess tremendous velocities, some approaching the velocity 
of hght (186,000 mi. /sec). So far as is known, the electron is 
the smallest division of matter, 

9. Kinetic Theory of Matter. — There is a well-substantiated 
theory that all molecules are in constant vibration. This is 
known as the kinetic theory of matter, 

10. Physical Change. — A physical change is one in which the 
substance does not lose its identity ^ such as planing a piece of 
steel; melting a quantity of tin; forming steam from water, 
etc. It is evident, in each case, that the new form assumed 
has the same molecular constitution as the original. 

11. Chemical Change. — A chemical change is one in which 
the substance loses its identity. If we burn a piece of coal, we 
find that it has been changed into ash and various gases, none 
of which resemble the original coal. Another example is the 
rusting of a nail. Rust is a combination of oxygen and iron 
and does not resemble the iron in any way. 

12. Force. — A force is anything that will cause a body to 
undergo a change of shape or a change in condition of motion. 
For example, a force may elongate, compress, twist or bend 
a body; it may cause a body at rest to move; it may cause a 
moving bod}^ to stop; it may increase or decrease the velocity 
of a body; or it may cause a change in the direction of the 
motion. 

13. Equilibrium. — A body at rest is in equilibrium, since its 
state of motion (zero) is constant; a flywheel rotating at a 
constant speed is in equilibrium, since its state of motion does 
not vary. Equilibrium means a balanced condition^ in which 
the state of motion does not change. 

14. General Properties of Matter. — There are certain 
characteristics that are common to all bodies. Such char- 
acteristics are called general properties. X few of the more 
important ones will be discussed. 

Porosity. — All matter is porous, that is, a space exists 
between the molecules of all bodies. No two adjacent mole- 
cules are ever in contact. It has been proved that gold is 



4 MECHANICS 

porous by forcing water through it under heavy pressure. 
A teaspoonful of sugar may be put into a cup of coffee without 
appreciably increasing the cubical contents of the cup. This 
is explained by the fact that the sugar molecules penetrate 
the inter-molecular spaces of the coffee. A simple experiment 
will demonstrate that water is porous. A long glass graduate 
is nearly filled with water and the graduation coinciding with 
the water line is noted. If a few drops of alcohol are intro- 
duced into the water and the mixture is thoroughly shaken, 
it will be seen that the water line has not risen. This is 
because the alcohol molecules have arranged themselves in 
the inter-molecular spaces of the water. 

Compressibility, — All bodies are compressible. Liquids are 
the least compressible and gases are the most compressible. 
Solids vary widely in this respect. Steel is difficult to com- 
press, while wood is relativel}^ easy to compress. When a 
body is compressed, the molecules are crowded closer together, 
causing the body to weigh less per unit volume. It is evident 
that compressibility is a direct consequence of porosity. 

Indestructibility. — All matter is indestructible. In burning 
a piece of coal, we do not destroy the matter of which the coal 
is composed. The coal is converted into ash and various gases, 
but the original matter still exists in other forms. 

Inertia. — Inertia^ commonly treated as a force, is a property 
of matter causing a body to resist any attempt to change its condi- 
tion of rest or motion. If a body is at rest, it resists any 
attempt to set it in motion. If a body is in motion, it resists 
any attempt to stop, increase or decrease its speed or to change 
the direction of its motion. 

Elasticity. — Whenever a body suffers a change of size or 
shape, it manifests a tendency to resume its original size 
or shape. This property is known as elasticity. Steel and 
copper are highl}^ elastic, while substances like rubber are 
less elastic. Even a mass of putty is slightly elastic. The 
elasticity of a body is measured by the amount of its resistance 
to a change of size or shape. 

Cohesion. — Cohesion enables like violecides to hold together. 



INTRODUCTORY 5 

It is greatest in solids, least in gases, and decreases with a rise 
in temperature. Tensile strength depends upon cohesion and 
is measured by the effort necessar}^ to rupture tensionally a 
body of unit cross-section. Strictly speaking, cohesion is a 
property of matter. We may think of it, however, as the 
force binding like molecules together. 

Adhesion. — Adhesion enables unlike molecules to hold together. 
Water clinging to a glass rod and paint clinging to a piece of 
wood are examples. Although adhesion is a property of 
matter, we may think of it as the force binding unlike mole- 
cules together. 

Mass. — Mass refers to the amount of matter a body contains. 
It is measured by the resistance that a body offers to a change 
of state of motion. The inertia of a body is proportional to its 
mass. Mass should not be confused with weight. The weight 
of a body may vary, but the mass will remain constant. A 
body weighing 10 lb. at the earth^s surface, will weigh practi- 
call}^ nothing at the earth's center. The amount of matter 
will be the same in each case. 

15. Specific Properties of Matter. — Specific properties are 
common to certain bodies only. They enable us to distinguish 
one substance from another. Following is a discussion of a 
few of the more important ones. 

Tenacity. — Tenacity is the relative measure of cohesion in 
various bodies. The tenacity of steel is very great; copper 
is less tenacious. A carbon steel casting may require as high 
as 80,000 lb. /in. 2 to rupture it tensionally. Alloy steel is 
even more tenacious. Rolled copper has a tensile strength of 
approximately 40,000 Ib./in.^ 

Hardness. — Hardness is the resistance a body offers to being 
scratched or worn by another. A body that will scratch another 
is said to be the harder of the two. The diamond is the 
hardest of all known substances. If a body is moving rapidly, 
it will cut into another body which is harder. The cutting 
abihty of an emery wheel depends upon the fact that the 
emer}'- is very hard and that the particles of emery are 
moving with great speed. 



6 MECHANICS 

Hardness should not be mistaken for hrittleness, Brittle- 
ness is aptness to break under the application of small forces. 
Vanadium steel is hard and tough, while glass is hard and 
brittle. Steel is made very hard and brittle by heating it to a 
red heat and then plunging it into a bath of water or oil. 
By reheating and cooling slowly, the steel will become soft 
and flexible. The latter process is called annealing. The 
desired hardness for cutting tools is brought about by temper- 
ing. The steel is reheated slowly until it assumes a certain 
color and is then plunged into the bath of oil or water Lathe 
tools are reheated to a pale yellow color (approx. 400°F.)., 
while saws are reheated to a dark blue color (approx. 600°F.). 

Ductility. — Ductility enables substances to be drawn out in 
the form of a wire. Platinum may be draw^n into a wire 
.00003 of an inch in diameter. Glass is so ductile that it may 
be drawn out into fine threads and the threads w^oven into 
cloth. 

Malleability. — A body which may be hammered, rolled or 
pressed into thin sheets is said to be malleable. It is possible 
to make gold leaf .000003 of an inch in thickness, in which 
case the gold is transparent. 

Questions and Problems 

1. Define mechanics, statics, kinetics, matter, energy. 

2. Distinguish clearly between the molecule, atom and electron. 

3. What is meant by the kinetic theory of matter? 

4. Define and give examples of a physical change. 
6. Define and give examples of a chemical change. 

6. What is a force? What effects may it produce? 

7. What is meant by equilibrium? Give examples. 

8. What is the difference between general and sjiecific properties of 
matter? 

9. Define and discuss: porosity, compressibility, indestructibility, 
inertia, elasticity, cohesion, adhesion, mass. 

10. Define and discuss: tenacity, hardness, brittleness, ductility, 
malleability. 

11. What is meant b}' annealing? Tempering? 



CHAPTER II 
MEASUREMENT AND MEASURING INSTRUMENTS 

SECTION I. MEASURES AND WEIGHTS 

16. The English System. — In the United States we still 
employ the so-called English system of measurement for every- 
day work. The yard is taken as the standard of length, the 
pound as the standard of mass (weight) and the second as the 
standard of time. In applied physics the foot {^i of a yard). 
the pound and the second are often used as fundamental 
units. A system of measurement employing the three latter 
units is called the foot-pound-second system. It is usually 
spoken of as the f.p.s. system. The English system is incon- 
venient and has given way almost entirely to the metric 
system for purely scientific work. At present there seems to 
be a tendency toward the adoption of the metric system in this 
country for all purposes. The coming generation should 
thoroughly master the metric system, so that there will be as 
little confusion as possible when the change occurs. 

17. The Metric System. — About the time of the French 
Revolution, the government of France appointed a commission 
to devise a system of weights and measures to replace the 
awkward system then in use. This resulted in the establish- 
ment of the metric system in France. It has since been made 
compulsory in most civilized countries, England and the 
United States being exceptions. The metric system is 
extremely easy to use on account of its simplicity and decimal 
scale. In scientific work the centimeter, gram and second 
are used as fundamental units. A system of measurement 
employing the three latter units is called the centimeter- 
gram-second system. It is usually referred to as the c.g.s. 
system. 

7 



8 



MECHANICS 



The 7neter (m.) was adopted as the standard of length in 
the metric system. It was intended to be MojOOOjOOO of the 
distance from the Equator to the North Pole, measured on the 
meridian of Paris. This distance was computed incorrectly, 
however, and the standard meter to-day is the distance 
between two transverse scratches on a bar of platinum- 
iridium at a temperature of 0°C. This bar is preserved in the 
archives of France and replicas have been sent to all civilized 
countries. The meter is 39.37 inches long. Our yard is 
officially defined as ^'^^H^osj of a meter. The meter is 
divided into 10 equal parts or decimeters; 100 equal parts or 
centimeters; and 1,000 equal parts or millimeters. The 
centimeter, due to its more convenient size, is universally used 



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UiUl 



3 4 5 

I Centimeters 



11,1] 1 1 1 ,1 [ 1 ,1 1 ,1 



IIIIMllllllll 



IMIIII 



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1,111 1 , 1 ,1 1 



1 1 1 1 1 1 1 1 1 1 I M I ' I 

Inches 

2 3 

Fig. 1. — Comparison of the centimeter and inch. 



1,111.1 1 ,1 



in scientific work. Figure 1 shows the relation of the centi- 
meter and inch. 

The metric standard of mass (weight) is the kilogram (Kg.), 
It was intended to be the mass of one cubic decimeter {dm.^) of 
pure water at a temperature of 4°(7. Due to a very slight error, 
this relation is not exactly true. It is close enough, however, 
for all practical purposes. The prototype kilogram is a 
cylinder of platinum-iridium kept in the archives of France, 
copies of which have been furnished to all civihzed countries. 
The gram (g.) or one thousandth of a kilogram is taken as the 
fundamental unit. The gram is divided into 10 equal parts or 
decigrams; 100 equal parts or centigrams; and 1,000 equal 
parts or milligrams. Our pound avoirdupois is equivalent to 
453.5924277 grams. For practical work, 454 grams is used. 
The kilogram is equivalent to about 2.2 pounds. Figure 2 
shows the practical relation of the pound and kilogram. 



MEASURING INSTRUMENTS 



9 



The unit of capacity in the metric system, both dry and wet, 
is the liter (1). It is equivalent to the volume of a cube one 
decimeter on a side and is approximately the same as a quart. 
The Hter is divided into 10 equal parts or deciliters; 100 equal 
parts or centiliters; and 1,000 equal parts or milliliters. For 
practical purposes, a liter of water may be considered to 




Fig. 2. — One kilogram weighs 2.2 pounds and just balances 1,000 cubic 

centimeters of water. 

weigh one kilogram. Hence one cubic centimeter of water 
weighs one gram. 

The unit of time in the metric system is the second. It is 
defined as ^6^400 of the mean solar day. 

The following tables will familiarize the student with the 
essential parts of the metric system. Heavy type is used for 
the more common units. 

MEASURES OF LENGTH 



10 millimeters (mm.) 

10 centimeters 

10 decimeters 

10 meters 

10 dekameters 

10 hectometers 

10 kilometers 



1 centimeter (cm.) 
1 decimeter (dm.) 
1 meter (m. ) 
1 dekameter (Dm.) 
1 hectometer (Hm.) 
1 kilometer (Km.) 
1 myriameter (Mm.) 



10 



MECHANICS 
MEASURES OF SURFACE 



100 square millimeters (mm. 2) 

100 square centimeters 

100 square decimeters 

100 square meters 

100 square dekameters 

100 square hectometers 



1 square centimeter (cm. 2) 
1 square decimeter (dm. 2) 
1 square meter (m.^) 
1 square dekameter (Dm. 2) 
1 square hectometer (Hm.^) 
1 square kilometer (Km. 2) 



MEASURES OF VOLUME 



1000 cubic miUimeters (mm.^; 
1000 cubic centimeters 
1000 cubic decimeters 



= 1 cubic centimeter (cm.') 
= 1 cubic decimeter (dm.') 
= 1 cubic meter (m.') 



MEASURES OF CAPACITY 



10 milliliters (ml.) 
10 centiliters 
10 deciliters 
10 liters 
10 dekaliters 
10 hectoliters 



1 centiliter (cl.) 
1 deciliter (dl.) 
1 liter (1) 
1 dekahter (Dl.) 
1 hectoliter (HI.) 
1 kiloHter (Kl.) 



MEASURES OF WEIGHT 



10 milligrams (mg.) 
10 centigrams 
10 decigrams 
10 grams 
10 dekagrams 
10 hectograms 
10 kilograms 
10 myriagrams 
10 quintals 



1 centigram (eg.) 

1 decigram (dg.) 

1 gram (g.) 

1 dekagram (Dg.) 

1 hectogram (Hg.) 

1 kilogram or kilo (Kg.) 

1 niyriagram (Mg.) 

1 quintal (Q) 

1 tonneau or metric ton (T.) 



The following; equivalents should be memorized by the pupil 
They are sufficiently accurate for all ordinar}^ work. 



1 meter = 39 . 37 inches (exact) 

1 inch = 2.54 centimeters (approx.) 

1 kilometer = .62 mile (api)rox.) 

1 kilogram = 2.2 pounds (approx.) 

1 liter = 1.06 liquid quarts (approx.) 



MEASURING INSTRUMENTS 11 

, QUESTIONS AND PROBLEMS 

1. Give a brief history of the metric system. 

2. Show why the metric system should be made compulsory in this 
country. 

3. What is the f.p.s. system of measurement? The c.g.s. system? 

4. Define: yard, foot, pound, meter, kilogram, gram, liter and second. 
6. Why must the prototype meter be kept at a temperature of 0°C. ? 

6. What is the weight in grams of 10 cm.^ of water? 

7. On July 2, 1901, the Cornell Varsity Crew at Poughkeepsie rowed 
4 mi. in 18 min., 533-^ sec, establishing the record for the course. What 
was the average speed in meters per sec? In kilometers per hr? 

8. The amateur record for the hundred yard dash is 9% sec. Express 
the speed in meters per sec. In kilometers per min. 

9. The muzzle velocity of a projectile was 2,000 ft. per sec. How 
many meters per hr. was this? 

10. An aviator ascended to a height of 30,000 ft. State the altitude 
in meters. 

11. Light travels with a velocity of 300,000 kilometers per sec. Give 
the equivalent velocity in feet per sec. 

12. If the barometer reads 760 millimeters, what will it read in inches? 

13. The wheel base of an automobile is 120 inches. What is it in 
centimeters? 

14. The diameter of the earth is about 8,000 miles. Express the 
diameter and circumference of the earth in kilometers. 

16. A fly wheel two meters in diameter rotates at the rate of 200 
r.p.m. What linear distance will a point on the circumference travel in 
4 minutes? 

16. An iron casting weighs 200 lb. What is its weight in kilograms? 
in metric tons? 

17. An aluminum block weighs 10 Kg. What is the weight in pounds? 

18. How many cubic centimeters are there in a granite block 2.5 
meters on a side? 

19. Compute the number of cubic inches in a cubic meter. 

20. A gasoline tank holds 15 gallons. Express its capacity in liters. 

21. A tank is 2 X 2 X 6 meters. How many grams of water will it 
hold? 

22. The drip vat of a steam power plant is 2 X 2 X 3 meters. How 
many cubic feet of drip will it hold? 

23. How many quarts are there in a cubic meter? 

SECTION II. MEASURING INSTRUMENTS 

18. The Steel Rule. — A 6 in. flexible steel rule, graduated in 
sixty-fourths of an inch on one side and millimeters on the 



12 



MECHANICS 



other, should be the personal property of each student. It 
is not expensive to buy, is practically indestructible, and may 
be carried in the pocket. 




Sfii* J »■« >. -4 THE USHSTARRETTCa ^ 

^iliTempered N9 4 1 athowmass u.s.a ' 2 



NQ 300 



ililililllililililililililililiiiiliiililililililifililil^ 



Fig. 3." — Steel rule. 



19. The Protractor. — Protractors are used to measure angles 
and are either adjustable or non-adjustable. One or more 
large adjustable protractors should be in every laboratory, 
and the student should own a small brass or steel protractor 




Fig. 4. — Ordinary form of protractor. 




^ ^ ^-^H^^ 



;t'» too'_co. --^ 



Fig, 5. — Adjustable protractor. 



of the non-adjustable type. The exact method of measuring 
an angle may be seen from a studj^ of the accompanying 
diagrams. 

20. The Ordinary Caliper. — Calipers, shown herewith, arc 
used in taking external and internal measurements when a 
high degree of accuracy is not necessary. They are either 
of the '^spring'' or '' firm-joint '' type. Outside and inside 



MEASURING INSTRUMENTS 



13 



calipers, of different capacities, should alwaj^s be available 
for laboratory work. 





Outside Inside 

Fig. 6. — Spring calipers. 





Outside Inside 

Fig. 7. — Firm-joint calipers. 



21. The Speed Indicator. — The speed indicator is an instru- 
ment for measuring the number of revolutions of shafts^nd 
spindles. The Starrett indicator shown here registers in either 



14 



MECHANICS 



direction up to 100. The rotating disc carries a raised knob 
which should coincide with the knob on the dial at the start. 
The disc is carried around by friction; hence it is an easy matter 
to set it back to zero. Each complete rotation of the disc 
indicates 100 r.p.m. of the shaft or spindle under test. In 




^ ^ _ r-i .Rubber 

Fig. 8. — Speed indicator. 

principle, the indicator is a worm and worm wheel. The 
student will be able to use the indicator without detailed 
instructions. 

22. The Slide Rule.— The slide rule is a device to facilitate 
the solution of numerical problems. Its accuracy is sufficient 
for most kinds of work. Problems involving multiplication, 
division, roots, powers, trigonometric functions, etc., may be 
solved with a great saving of time. Students will find the 



'■'■'^"' .■ 



.=;ji 




•f<4<«ii*i 



53, 



Fig. 9. — Polyphase slide rule. 

sHde rule of great convenience for both home work and class 
work. It is comparatively expensive, yet it is hoped that a 
majority of students will be able to own one. The Polyphase 
rule shown in the diagram, has proven very satisfactory. 
Complete and detailed instructions are furnished by the 
manufacturer with each rule. 

23. The Micrometer Caliper.— The micrometer caliper is 
used to take measurements when very fine readings are 
desired. The type shown in Fig. 10, is for external measure- 
ments. Micrometer calipers may be procured in various 



MEASURING INSTRUMENTS 



15 



sizes, some reading as high as 24 in. Ordinarily, they read 
only to thousandths of an inch, but by careful estimation of 
scale divisions, finer readings are obtainable. They are also 
graduated metrically with corresponding accuracy. The 




Fig. 10. — Micrometer caliper. 

one described here will read accurately to a thousandth of an 
inch. 

The micrometer caliper consists of the following essential 
parts: the frame {F); the anvil (A); the spindle {S); the 
sleeve (H); and the thimble (T). 

The sleeve is graduated in tenths of an inch. The grabua- 
tions are labelled 1, 2, 3, etc. Each numbered division is 
sub-divided into four equal parts or fortieths of an inch. The 
spindle has forty threads to the inch and rotates in a fixed nut. 



0" 




t 3 4 


^ 


5 

20 




1 


1 1 


II III III 


^ 










(a) 
4^^o .o. ^4'f 

10^ ^0/000 10 
. 400^. 000 H 000 -. 400 



(b) 

JO 40 1000 1000 
.?00i-.0S0^.0ll=.261 



Fig. 11. — Reading a micrometer caliper. 

Hence one full rotation of the thim.ble will cause the spindle to 
move one fortieth of an inch to the left or right. Since the 
bevelled edge of the thimble is divided into twenty-five equal 
parts, a movement of one scale division on the bevel will cause 
a spindle movement of one twenty-fifth of one fortieth or one 
thousandth of an inch. 



16 



MECHANICS 



In using the micrometer caliper, the body to be measured 
should be placed in the jaws of the instrument and the thimble 
rotated until light contact is made. If the caliper is forced^ 
inaccurate readings will result. A ratchet device, automatically 
stopping the spindle at the proper pressure, is often provided. 
The '^spindle'' reading plus the 'Hhimble^^ reading will give 
the dimension sought. 

The diagrams in Fig. 11 represent the relative positions of 
the spindle and thimble for the readings indicated. The 
student should study these diagrams very carefully. 

24. The Vernier Caliper. — The vernier caliper, like the 
micrometer caliper, is used when very fine measurements are 
required. The one shown below will read down to one 




iHlfliilMiWili^iUiri^irtililiiyi 




^£LS 



Fig. 12. — Vernier caliper. 



thousandth of an inch. It is graduated in thousandths of an 
inch on the front and sixty-fourths of an inch on the back. 
Only the front will be discussed. 

Figure 12 shows the general construction of a standard 
vernier caliper. It consists of the following parts: the bea7n 
(A); the slide (B); the hinding screws (C and D); the adjusting 
screw (E); and the jaivs (F and G). 

The beam (Fig. 13) is graduated in inches and is numbered 
1. 2, 3, etc., large numbers being used. The inches are 
divided into ten equal parts or tenths of an inch and are num- 
bered 1, 2, 3, etc., small numbers being used. The tenths are 
further divided into four equal parts or fortieths of an inch and 
are not numbered. The smallest division on the beam is, 
therefore, one fortieth (.025) of an inch. 

The slide (Fig. 13) is divided into tweniy-jive equal parts 
and is numbered 5, 10, 15, 20 and 25. Each numbered part 



MEASURING INSTRUMENTS 



17 



is sub-divided into five equal parts without numbers. There 
are in all, then, twenty-five small divisions on the slide. 

By reference to Fig. 14, we see that twenty-four small 
divisions on the beam correspond in length to twenty-five 
small divisions on the slide. Hence a small division on the 
slide = 2^5 of a small division on the beam = 2^^^ of 
.025 in. = .024 in. Further, we see that the first graduation on 





|^il■l]lHlllllllHlHlnlllMllllllmllUlll»■ll 

2 



45 G 7 8 9 



TTTITTTT 



5 10 15 30 25 



■timmi.im. 



Fig. 13. — Vernier caliper, view of beam and slide.^ 



Sma// 6rac/c/af/b/7S = .025 




BEAM 



SLIDE 



Srnal! Gwoluafions= .034" 
Fig. 14. — Vernier caliper, comparison of small graduations on beam and 

slide. 



the slide differs from the first graduation on the heain by .001 in. ; 
that the second graduation on the slide differs from the second 
graduation on the beam by .002 in., etc. 

The following instructions should be observed in manipulat- 
ing the vernier. The instructions are intended only for 
external measurements and refer to Fig. 12. After moving 
the slide over until both jaws nearly touch the object, the 
binding screw (D) should be screwed down. The fine adjust- 
ment (E) should be rotated until each jaw touches the object 



18 



MECHANICS 



lightly. After screwing down the binding post (C) so as to 
lock the jaws, the reading may be taken. 

To illustrate the reading of a vernier caliper, let us take an 
example and refer to Fig. 15. In measuring the diameter of a 
steel rod it was found that the twenty-ninth small division on 
the beam exactly lined up with the eighteenth small division 
on the slide. It is evident that the diameter of the rod is the 
linear distance represented by the twenty-nine small divisions 
on the beam less the linear distance represented by the eighteen 
small divisions on the sHde, viz., .725 in. - .432 in. = .293 in. 

BEAM 




SLIDE 
Fig. 15. — Vernier caliper, showing coincidence of graduations X and Xi. 

Since the zero mark on the slide (Fig. 15) extends somewhat 
beyond the .275 in. mark on the beam, it will be seen that the 
diameter of the rod is also .275 in. plus the distance to zero on 
the slide. As the points of coincidence (X and X^) are eighteen 
slide divisions beyond zero on the slide, then zero on the slide 
must extend .018 in. beyond the .275 in. mark on the beam. 
Hence the correct reading will be .275 in. + .018 in. = .293 in., 
the same as determined in the first method. The method last 
described should always be used as it is more direct. 

To read the vernier caliper we may adopt the following 
rule : Note and record the first small division on the beam to the 
left of zero on the slide. Add to this .001 in. for each small 
slide division up to the point where the beam and slide divisions 

coincide. 

OUeSTIONS AND PROBLEMS 

1. Descril)c two types of calipers used for approximate work. 

2. For what is the micrometer cahper used? Describe its construc- 
tion and operation. 



MEASURING INSTRUMENTS 19 

3. For what is the vernier eahper used? Describe its construction 
and operation. 

4. The following readings were obtained in reading a micrometer 
caliper: .240 in.; 1.361 in.; and 2.546 in. What is the sleeve and 
thimble reading in each case? 

5. If the above readings were obtained with a vernier caliper, state 
in each case the point at which the beam and slide division coincided. 



CHAPTER III 



ELEMENTARY TRIGONOMETRY 



25. Trigonometry Defined — Trigonometry deals with the 
relations which the sides and angles of a right-angled triangle 
bear to each other. An elementary knowledge of trigonometry 
facilitates the solution of many numerical problems. Hence it 
is very important for the student to gain a working knowledge 
of the subject. 

26. Trigonometric Functions. — A trigonometric function is 

the ratio which one side of a right-angled 
triangle bears to another. There are six 
such relations. They are given here with 
respect to angle 6. The first three 
functions are sufficient for ordinary 

Fig. 10. purposes. 




AB 
AC 
AB 
BC 
AC 
AC 
BC 
AB 
AC 
AB 
BC 



_ side opposite __ 



hypotenuse 
side adjacent 

hypotenuse 
side opposite 
side adjacent 
side adjacent 
side opposite 

hypotenuse 
side adjacent 

hypotenuse 
side opposite 



= sine (sin) of angle 0, 
= cosine (cos) of angle d. 
= tangent {tan) of angle 6. 
= cotangent {cot) of angle 6. 
= secant {sec) of amjle 6. 
= cosecant {cosec) of angle 6. 



27. Trigonometric Solutions. — Let us illustrate the solution 
of problem.^ by the use of the sine, cosine and tangent. Refer- 

20 




ELEMENTARY TRIGONOMETRY ' 21 

ring to Fig. 17, we have a right angled-triangle. If the side 
BC is5 inches in length, by actual measurement the other side 
will be 8.66 inches and the hypotenuse 10 inches. As long as 
the angles are kept constant, even if the lengths of the sides 
are changed, there will always exist the same numerical ratio 
between the lengths. 

Solutions. — 1 Suppose it is desired to find 
BCy knowing the angle BAC and the hypotenuse 

AB. Solution: BC/AB = sin BAC, BC = AB 
sin BAC, Referring to the trigonometric tables 
in the appendix, we find that the sine of angle 
BAC (30°) is .500. Hence, by substitution, 
2^C = 10 X .500 = 5. 

2. Suppose it is desired to find ACy knowing angle BAC and the 
hypotenuse AB, Solution: AC/AB = cos BAC, AC = AB cos BAC. 
By substitution, AC = 10 X .866 = 8.66. 

3. Suppose it is desired to find 5C, knowing angle BAC and the side 

AC. Solution: BC/AC = tan BAC, BC = AC tan BAC. By substi- 
tution, BC = 8.66 X .577 = 4.996. (Since the tables in the appendix 
are to three decimal places only, we get 4.996 instead of 5 as expected. 
Three place tables are accurate enough for ordinary purposes.) 

4. Suppose it is desired to find angle BAC^ knowing sides AB and BC, 
Solution: Sin BAC = BC/AB. Sin BAC = Ko = .500. By reference 
to the trigonometric tables, it is found that the angle to which .500 
corresponds is 30°. 

Questions and Problems 

1. With what does trigonometry deal? Why is it important in 
mechanics? 

2. What is a trigonometric function? 

3. Name and define the six trigonometric ratios. 

4. A guy wire attached to the top of a telephone pole is secured to the 
ground by means of a "dead man'' and makes an angle of 60° with the 
ground. The wire enters the ground 18 ft. from the bottom of the pole. 
(a) How long is the wire? (b) How high is the pole? 

5. A flag pole rope, when pulled 20 ft. away from the bottom of the 
pole, just touches the ground. If the rope pulley is 40 ft. above the 
ground, (a) how long is the rope? (5) What angle does it make with 
the ground? 

6. The Woolworth building is 792 ft. high. An imaginary line, at an 
angle of 60° to the vertical, is drawn from the top of the tower to the 
ground. Assuming no change in altitude and no curvature of the earth, 



22 MECHANICS 

(a) how far horizontalh' from a point directly under the tower would the 
line strike the ground? (b) How long would the line be? 

7. A ladder 30 ft. long is placed against a house at an angle of 48° 
to the horizontal, (a) How high is the top of the ladder above the 
ground? (6) How far is the bottom of the ladder from the house? 

8. A shadow cast by the top of a monument 30 ft. high is 12 ft. long. 
What is the angle of the sun with reference to the ground? 

9. A captive balloon 200 meters high is driven by the wind until the 
tie rope is at an angle of 80° to the ground. What is the second height 
of the balloon ? 

10. From a light house 50 ft. high, two rocks are seen in a direct line 
with the observer. If the rocks make angles of 60° and 70° respectivel}' 
with the vertical at the top of the light house, how far apart are the rocks? 



CHAPTER IV 
GRAVITATION AND GRAVITY 

28. Universal Gravitation. — It has been found that there is 
a mutual attraction existing between all bodies in the universe. 
For instance, there is an attraction between the masses of the 
earth and the moon, although they are separated by a distance 
of over 240,000 miles. The periodic rising and falling of the 
ocean waters, known as tides, is ascribed 

largely to the gravitational effect of the r\ ^ r\ 

moon. There is also an attraction ^ ^ 

between the earth and the sun, the 

earth and the stars, etc. The effect is 

slight, however, due to the tremendous A g 

distances between them. Cy "^ "(O 

29. The Law of Universal Gravita- 2m 2m 
tion. — A study of gravitation led to an 

enunciation by Sir Isaac Newton^ of 

^ B 

the following law : ^->^ ^-^ 

Between every two particles of matter v^ \^ 

in the universe there exists a mutual ^^ ^^ 

attraction which varies directly as the , ^^^\ i^— attractive 

*^ force between A and B in 

product of the masses and inversely as middle diagram is four 
the square of the interveninq distance, times as great as m upper 

. nil ^^^ lower diagrams. 

The above law is called the law of 
universal gravitation and will be readily understood by refer- 
ence to Fig. 18. A and B (upper diagram) represent two 
bodies with masses of M and m respectively and separated 

iSir Isaac Newton (1642-1727). Born in England. Buried in 
Westminster Abbey. Famous as a physicist, mathematician and 
astronomer. Professor of mathematics at Cambridge, member of 
Parliament and Master of the Mint. Responsible for the law of gravi- 
tation, the three laws of motion which beal> his name, the bionomial 
theorem, the calculus and many discoveries in light. 

23 



24 MECHANICS 

by a distance of d. The product of the masses is Mm, If 
the masses are doubled (middle diagram), their product will 
be 4 Mm and the mutual gravitational force will be increased 
4 times. If, in addition, the distance between A and B is 
doubled (lower diagram), the gravitational attraction between 
them will be reduced to 3^^ its former value. The attraction 
in the upper and lower diagrams is therefore identical. If 
the distance is tripled and the masses doubled in the upper 
diagram, the attractive force will be ^^ of its original intensity. 

30. Gravity. — The term gravity refers to gravitation in a 
restricted sense. Specifically, it means the mutual 
attraction between the earth and bodies on or near the 
earth^s surface. Thus the force which impels bodies 
toward the earth is called the force of gravity. 

The force of gravity tends to pull all objects towards 
the earth's center of mass. This point is approxi- 
mately at the geometrical center of the earth (4,000 
miles below the surface). The direction of the pull is 
conveniently determined by means of a plumb line. 
Ordinarily the plumb line points directl}^ toward the 
earth's center of mass. It will be deflected slightly, 

f however, if brought near a large mountain. A deflec- 
tion will also take place if the cord is too long, due to 
Fi(.. 19. the rotation of the earth. Ordinarily, deflections are 
line^ ^ ^^^ slight to havc any serious effects and, for practical 
purposes, we may assume that the plumb line, if 
extended, would pass through the center of the earth. We 
say that the plumb line assumes a vertical position. Any line 
drawn perpendicular to the plumb line is said to be horizontal, 

31. Weight. — The mass or quantity of matter possessed by 
a body remains constant, but the iveight is subject to variation. 
The weight of a body is the mutual attraction existing between 
the body and the earth. For example, if a man weighs 175 lb. 
at a certain place on the earth, the earth is pulling the man 
toward its center with a force of 175 lb.; likewise, the man is 
exerting a pull of 175 lb. upon the earth. The maximum 
weight is at the earth's surface. 



GRAVITATION AND GRAVITY 25 

32. Bodies Above and Below the Earth's Surface. — Frora 
the law of universal gravitation, it follows that the weight of 
a body above the earth^s surface will be inversely proportional 
to ihe square of its distance from the center of the earth. 
Algebraically : 

W :w ::d^ : Z)^ 

Suppose we wish to know how much a body weighing 1,000 
lb. at the earth^s surface will weigh 1,000 miles above the 
surface. Employing the above proportion we have: 

1,000 :w :: 5,000^ : 4,0002, from which w = 640 lb. 

A body below the earth's surface will lose in weight due to 
the fact that the particles above the body exert a contrary 
attraction to that of the particles below. The resultant force 
acting on the body will be the difference between the two 
attractions. 

As the polar diameter of the earth is less than the equatorial 
diameter, it follows that gravity is greater at the poles than 
at the equator. The effect of gravity is also lessened at the 
equator due to the centrifugal force set up b}^ the earth's 
rotation. If the earth were to rotate 17 times as fast as it 
does, the force of gravity at the equator would be neutralized. 

33. The Acceleration of Gravity. — Since the force of gravity 
varies from place to place, the acceleration imparted to 
falling bodies will vary slightly. A bod}^ at the equator will 
fall with less velocity than a body at the poles. Also, if two 
bodies fall an equal distance but strike the ground at different 
altitudes, the body striking at the lower altitude will have the 
greater velocity. The acceleration of gravity is the velocity a 
freely falling body will gain during each second of fall and will 
be written hereafter as ^'^.'' At the latitude of New York 
City (40.73°N.), the acceleration of gravity is 32.16 ft. per 
second per second or 980 cm. per second per second. 

34. Center of Gravity. — Since all bodies are composed of 
small particles, it naturally follows that each particle is acted 



26 



MECHANICS 



upon b}^ gravit}^ individually. Thus we ma}^ consider a 
body as acted upon by parallel forces, the sum of which con- 
stitutes the weight of the body. For convenience, we may 
assume that all of the separate forces may be replaced by a 
single force having the same effect as the joint action of the 
separate forces. The point at which the resultant force 
would be applied is called the center of gravity or center of 
mass of the bod3% The center of gravity is the point at which 
the iveight of a body may he considered concentrated. If a body 
is suspended at its center of gravity, there will be no tendency 
toward rotation. The weight of a body is always taken as a 
separate force acting downward at the center of gravity. 



35. Determination of the Center of 
Gravity. — In regular, homogeneous 
bodies, the center of gravity may be 
found geometrically. The center of 



CG 





Fig. 20. — Center of gravitj^ determined 
geometrically. 




(b) 

Fig. 21.— Center of 
gravity determined ex- 
perimentally. 



gravity of a uniform lever is at its middle point; of a rectangle, 
the intersection of the diagonals; of a circle, the center; of 
a triangle, the intersection of the medians, etc. (See Fig. 20.) 
If the body is irregular in shape, the center of gravity may be 
determined experimentally. For example, suppose we have a 
piece of sheet zinc as shown in Fig. 21(a). The zinc is freely 
suspended from hole ''a'' and a plumb line dropped from the 
point of suspension. The center of gravity will be along 
this line which is marked. The zinc is then suspended 
from hole '^6^^ and the previous operation is repeated. The 



GRAVITATION AND GRAVITY 27 

center of gravity will be at the intersection of the two posi- 
tions assumed by the plumb line. If a hole is drilled at the 
point of intersection, the sheet will remain fixed in any posi- 
tion. The center of gravity of levers, rods, etc., may be found 
by balancing over a knife edge (Fig. 21 (6) ). More compli- 
cated determinations of the center of gravity are omitted 
as being beyond the scope of this book. 

36. Equilibrium. — A body is said to be in equilibrium with 
respect to gravity when a vertical line from its center of 
gravity passes through the supporting base. If the vertical 
line passes outside of the supporting base, a rotation results 
and the body is overturned. There are three kinds of equi- 
librium; (1) stable; (2) unstable; and (3) neutraL 




Fig. 22. — Bodies A, B and C are respectively in stable, unstable and neutral 

equilibrium. 

The three kinds of equilibrium are illustrated in Fig. 22. 
A J B and C are metal disks with their centers of gravity as 
shown. If A is freely suspended at point Pi, it is said to be in 
stable equilibrium. If displaced, the body tends to resume 
its former position, due to the fact that any rotation necessi- 
tates a raising of its center of gravity. Since B is suspended 
at P2, any displacement will lower the center of gravity and 
the body will not return to its first position. B is therefore 
in unstable equilibrium. C is suspended at P3. Since this 
point coincides with the center of gravity, it is evident that a 
displacing force will neither raise nor lower the center of 
gravity and that the body will stay wherever it is put. C is 
therefore in neutral equilibrium. 



28 



MECHANICS 



37. Stability. — A body is said to possess stability if it is 

difficult to overturn. Stability depends 
upon the size of the supporting base 
and the location of the center of gravity. 
A large supporting base and a low center 
of gravity give a body great stability. A 
racing yacht must be well ballasted to 
keep it from overturning. Cargo ships 
are always ballasted when crossing the 
ocean with no cargo aboard. The 
center of gravity is thus kept low. 
Tall chimneys are not highly stable, 
Fig. 23.— Body A-B-C-D ^g ^j^^ supporting base is relatively 

has 5 ft.-lb. of stability. „ ,^\ ^ ^ . ^ -^ 

small and the center of gravity rela- 
tively high. The stability of a body is measured by the 
amount of work necessary to overturn the body. It is found by 
multiplying the weight of the body (concentrated at the 
center of gravity) by the vertical distance through which the 
center of gravity is raised. Thus in Fig. 23, the body ABCD 
has a stability of 5 foot-pounds. 




Questions and Problems 



1. What is meant by universal gravitation? 

2. State and illustrate the law of universal gravitation. 

3. Distinguish between gravity and weight. 

4. For what is a plumb line used? Does it always point to the center 
of mass of the earth? Explain. 

6. Why does a body weigh most at the earth's surface? 

6. State the formula for finding the weight of bodies alx)ve the 
earth's surface. 

7. State two reasons why a body weighs less at the equator than at 
the poles. 

8. What is meant by the acceleration of gravity? What factors 
affect it? What is the acceleration at New York City? 

9. Explain fully what is meant by the center of gravity. 

10. State various methods of determining the center of gravity. 

11. Define equilibrium. Describe the three kinds of equilibrium. 

12. What is stability? How is it measured? Give examples of high 
and low stability. 



GRAVITATION AND GRAVITY 29 

13. A body weighs 50 lb. at the earth^s surface. How much will it 
weigh 5,000 miles above the surface? 

14. A body weighs 50 lb. 10,000 ft. above the earth's surface. How 
much will it weigh at the surface? 

15. A body weighs 100 lb. at the earth's surface. How high above the 
surface would it have to be in order to weigh 40 lb. ? 

16. A body 4 X 3 X 2 ft. weighs 200 lb. Compute its stability as it 
rests on each face. 



CHAPTER V 
FORCES 

38. Force. — Before attempting a definition of force, let us 
examine the various effects which a force may produce. It is 
evident, from common experience, that a force may: 

(a) Cause a body at rest to move, 
{h) Bring a moving body to rest, 
(c) Cause a body to increase or decrease its speed, 
{(i) Alter the course of a moving body, 

(e) Cause a change in the shape of the body by stretching, compressing, 
bending, twisting it, etc, 

(/) Cause certain combinations of the above effects. 

No better or simpler illustrations of the efifects of forces 
can be had than from our national game, — baseball. Let us 
assume that the teams are on the field and that the batter is 
^'up.^' The ball, which is at rest (zero motion), will have 
velocity imparted to it as the pitcher makes his delivery. 
If the batter swings and misses, the ball will settle to rest in 
the catcher's mitt. In this case, the ball is stopped both by 
the muscular effort of the catcher and the inertia of the mitt. 
On the second delivery, the batter may drive a ball served to 
him slowly with such tremendous velocity that it will clear 
the stand or, if the delivery is fast, bunt it slowly into the 
infield, in either case changing both the amount and direction 
of the original motion. In addition, the impact of the ball 
and bat tends to change the shape of the ball. 

From the above discussion, it is evident that a force is 
anything tending to cause a change in the amount or direction of 
the motion possessed by a body or a change in the shape of the 
body itself, 

39. Action and Reaction. — Whenever one body acts upon 
another body, the body acted upon exerts an equal and 

30 



Bo 



ok 



{fable 



FORCES 31 

opposite effect commonly called a reaction. The following 
illustrations will make clear the difference between action and 
reaction. 

1. Suppose we have a book weighing 2 lb. resting upon a table as in Fig. 
24. It will be seen that gravity pulls 

the book against the table with a force p^ 

of 2 lb. It will be seen, also, that the 

table pushes up against the book with a 

force of 2 lb.; otherwise the book would 

move down through the table. Gravity _ 

is the action, and the upward force 

exerted by the table is the reaction. ^# 

2. A horse is tied to a post by means Fig. 24. — Action and reaction, 
of a halter. If the horse pulls horizon- 
tally with a force of 200 lb., the post must exert a force of 200 lb. 
in the opposite direction. The horse causes the action and the post the 
reaction. 

3. Two boys are pulling at the opposite ends of a rope. If one boy 
exerts a force of 40 lb., the other boy must exert a force of 40 lb. in the 
opposite direction. In this case, either boy may be considered as causing 
the action. 

4. An automobile is being towed over a level stretch at uniform speed. 
The force exerted by the tow rope is the action and friction is the reaction. 

5. If the automobile in (4) is being towed at increasing speed, the force 
exerted by the tow line is the action, and friction plus inertia is the 
reaction. 

6. If the automobile in (5) is being towed up an incline, the force 
exerted by the tow line is the action; and friction plus inertia plus the 
the effect of gravity is the reaction. 

40. Measurement of a Force. — We have seen that a force 
tends to change the amount or direction of the motion pos- 
sessed by a body or effect a change in the shape of the body. 
It will be clear, then, that the amount of change produced will 
determine the magnitude of the force. 

A force may be determined in magnitude: 

(a) By the velocity it will impart to a given body in a given 
time. 

(b^ By the amount of change of shape that it will cause a 
body to undergo. 

41. Graphical Representation of Forces. — In order to 
represent a force graphically, three factors must be considered: 



32 MECHANICS 

(1) the magnitude of the force; (2) the direction in which the 
force acts; and (3) the point at which the force is appKed to 
the body. This is readily done, in a quantitative way, by 
the use of an arrow drawn to a suitable scale. 

For example, suppose a body B (Fig. 25) to be acted upon 
by a pulling force of 100 lb. (Fi) acting west and another 
pulling force of 200 lb. (F2) acting east. Assuming that 1 in. 
= 200 lb., it is necessary only to draw a line 3^^ in. long due 
west from 5, and another line 1 in. long due east from B 

Scale . /"=P00^ Scale • / "=300"^ 
F,< 5 >^ yr ^ ^ 

Fig. 25. — Forces represented graphi- Fig. 26. — Forces represented graphi- 

cally, cally. 

The arrows show the direction of the forces, also that the body 
B is being subjected to tension, tending to pull apart the 
particles of which the body is composed. 

In case Fi and F2 act as pushing forces, the arrows would be 
placed as shown in Fig. 26. The body B is now under com- 
pression, tending to crowd the particles of which it is com- 
posed closer together. 

42. Moment of a Force. — We have seen that forces have a 
tendency to produce motion. Under certain conditions, the 

n_ ^^ motion may take the form of 
^ rotation about a fixed point 



AL 



24'^. _> 



H- — /2 - 



Fi( 



{fulcrum or axis of rotation), 
]3 For example, suppose AB 
(Fig. 27) to be a rigid bar 
without weight, arranged to 
^^^^ turn about a pin at the 

fulcrum (/) and acted upon 
by the forces Fi and F2. It is evident that Fi tends to 
produce a clockwise rotation and F2 a counter-clockwise 
rotation. Each force, then, tends to set up an opposite rota- 
tion about /. 

The measure of the tendency of a force to produce rotation 
about a given point is called the moment of that force. 



FORCES 



33 



H 



^^ 




"7" 



\B 



y 




Fig. 28. 



Referring again to Fig. 27: 

The moment of Fi = 20 X 12 = 240 pound-inches. 

The moment of F2 = 10 X 24 = 240 pound-inches. 

The rotative effect of each force is the same and the bar AB 
will be at rest. Thus a small 
force with a long moment 
arm may produce the same 
results as a large force with 
a small moment arm. 

The moment arm of a force 
is the perpendicular distance 
from the line in which the 
force acts to the fulcrum or 
point of rotation. The 
Moment^ of a Force = the 
Force X the Moment Arm. 

In the case of forces acting at an angle, it is necessary to 
find the moment arm as shown in Fig. 28. Thus the moment 
arms of Fi and F2 are di and ^2 respectively. A determination 
of the perpendicular distances often necessitates an extension 
of the force lines. The extensions should always be dotted 
lines. 

Questions and Problems 

1. State the various effects which a force may produce. 

2. From your own observations, give practical illustrations of the 
above. 

3. Define force. How may a force be represented graphically? 

4. Give examples of tension and compression. 

5. State two ways by which a force may be measured. 

6. Under what conditions will a force produce rotation? 

7. What is meant by the moment of a force? 

8. Define moment arm; fulcrum; torque. 

9. State clearly how you would find the moment of a certain force. 

10. Explain why forces never exist singly. 

11. Distinguish carefully between action and reaction, giving simple 
illustrations. 

12. Indicate the actions and reactions in the following cases: 

^ For pulleys, etc., the moment is often called torque. 
3 



34 MECHANICS 

(a) A team of horses towing an automobile on the level. 

(b) A team of horses towing an automobile up a hill. 

(c) A team of horses towdng an automobile up a hill at increasing 
speed. 

(d) Firing a gun. 

(e) A train coasting into a station. 

(/) A Hudson River steamboat proceeding north at increasing speed; 
the same boat going down stream at increasing speed. 
(g) The roof truss (Fig. 63). Use selected points. 



CHAPTER VI 
MOTION 

43. Motion. — Motion is the change of position which a body 
undergoes with reference to some given point or object and is 
always brought about by the apphcation of force. Rest 
means a permanence of position with reference to some given 
point or object. Rest is simply a statement of zero velocity, 
indicating lack of change of position with respect to some point 
or object. 

44. Absolute and Relative Motion. — If we wish a highly 
scientific statement of motion, we must record the successive 
changes with respect to some ideal point fixed in space. The 
motion is then absolute. 

Ordinarily we think of motion with respect to some nearby 
object (or objects), in which case the motion is relative. In 
defining the speed of an automobile as 30 m.p.h., the ground 
is taken as a basis of comparison. Actually both the ground 
and automobile are in motion on account of the earth's rota- 
tion and revolution. Thirty m.p.h. simply means that the 
speed of the automobile in a given direction exceeds that of the 
earth in the same direction by 30 miles per hour. 

Two express trains, traveUing at the rate of 60 m.p.h. in 
the same direction, have a velocity of zero with respect to 
each other, and a velocity of 60 m.p.h. each with respect to the 
earth. The same trains, travelling in opposite directions, 
have a velocity of 60 m.p.h. with respect to the earth and a 
velocity of 120 m.p.h. with respect to each other. 

A man walking forward in a submerging submarine will 
have various velocities. He will have a velocity with respect 
to the land, the water, members of the crew, a torpedo just 
discharged, etc. It is evident, from the foregoing illustra- 

35 



36 MECHANICS 

tions, that for practical purposes all motion is relative; that is, 
to define the motion of any body, we must refer it to some 
nearby body as a basis of comparison. 

45. Translatory and Rotary Motion. — In order to have 
translatory motion {motion of translation), every particle of 
which a rigid body is composed must have the same hnear 
velocity at each instant as every other particle. It is evident 
that each particle will travel in a path either parallel to, or 
coincident with, that of its neighbor. Furthermore, a straight 
line connecting any two particles will always keep the same 
direction. Translatory motion may be curvilinear (in a 
curved line) or rectilinear (in a straight line). It may also be 
uniform or variable. The motion of an engine piston and a 
pile driver are good examples of rectilinear-translatory motion. 
A body projected horizontally from a tower without rotation 
is a good example of curvilinear-translatory motion. 

In rotary motion {motion of rotation), each particle travels 
at the same angular velocity as the other, but particles 
farthest from the axis of rotation will possess the greatest 
linear velocity. It will be seen that each particle will describe 
a course either parallel to, or coincident with, that of its 
neighbor; and that only particles equally distant from the 
axis will travel with the same linear velocity. The motion of 
a flywheel and an airplane propeller are common examples. 

46. Accelerated Motion. — Acceleration is the rate at which 
the velocity of a body is increased or decreased. For example, if 
an automobile, starting from rest, increases its velocity 2 ft. 
every second for 10 consecutive seconds, the acceleration is 
then 2 ft. per second per second (written 2 ft./sec.^). This 
simply means that the velocity increases at the rate of 2 ft./ 
sec. during every second. The same acceleration may be 
expressed as 120 ft./min./sec. In the above illustration the 
acceleration is uniform and positivCj that is to say, the change 
of velocity is constant and causes an increase in the speed of 
the automobile. When a body is so acted upon by an accelera- 
tion that a decrease in velocity results, the acceleration is 
negative. For example, if a projectile is shot vertically upward 



MOTION 37 

with a velocity say of 1,000 ft. per sec, the acceleration of 
gravity causes the body to lose 32.16 ft. per second during each 
second and will eventually stop its flight. The acceleration 
of gravity is negative for rising bodies and positive for falling 
bodies. 

47. Speed and Velocity, — Speed is the rate at which a body 
moves. It concerns itself with displacement and time only, the 
direction not being taken into consideration. An express 
train travelling at the rate of 60 m.p.h. is said to have a speed 
of 60 m.p.h. In discussing the velocity of the train, we must 
include another element, — direction. If the train in proceed- 
ing due north, it has then a velocity of 60 m.p.h. due north. 

In case the velocity of a body is variable^ it is expressed as if 
the velocity of the body were to become uniform at the 
particular instant. This is called instantaneous velocity. A 
body with an instantaneous velocity of 10 m.p.h. will not 
necessarily cover 10 miles in an hour. The distance covered 
will depend on the average velocity for the given time. 

The velocity (v) at the end of any second for a uniformly 
accelerated body starting from rest is equal to the accelera- 
tion (a) multiplied by the time (t). Thus the velocity of a 
freely falhng body at the end of the first 5 seconds will be 
32.16 X 5 or 160.80 ft. /sec. 

V = at 

If a body is uniformly accelerated, the average velocity 
(v av.) may be found by adding the initial velocity {v in.) 
to the final velocity {v fin.) and dividing by 2. To illustrate, 
suppose a body falls from a state of rest. Since the accelera- 
tion of gravity is 32.16 ft./sec.2, the velocity at the end of 3 
seconds will be 96.48 ft. /sec. The average velocity for the 
3 seconds will be (0+96.48)/2 or 48.24 ft./sec. 

V in. + V fin. 
V av. = — - — 

In case the velocity of a body is uniform, the total distance 
(S) covered in a number of seconds is equal to the uniform 
velocity (v un.) multiphed by the time (t) in seconds. For 



38 MECHANICS 

example, if a bod}" has a uniform velocit}^ of 1,000 ft. sec, in 
10 seconds it will travel, 1,000 X 10 or 10,000 ft. 

S = V 1171. X t 

If the velocity of a body is uniformly accelerated, the total 
distance {S) covered in a number of seconds (t) will equal the 
average velocity (v av.) multiplied by the time. Accordingly, 
a freely falling body will cover 144.72 ft. during the first 
3 seconds of fall. The average velocity for the 3 seconds will 
be (0 + 96.48) /2 or 48.24 ft./sec. 48.24 X 3 = 144.72 ft. 

S = V av. X t 

48. Momentum. — The quantity of motion possessed by a 
moving body is called momentum. Momentum is measured 
by the product of the mass and velocity of the hody. 

Momentum = Mv 

There is no recognized unit of momentum in the English 
system of measurement. For practical purposes we ma}" 
assume the unit to be a mass of one pound moving at the rate 
of 1 ft. per second. For example, a 10,000 ton ship with a 
speed of 20 ft./sec. is said to have 400,000,000 f.p.s. units of 
momentum (10,000 X 2,000 X 20). 

In the c.g.s. system the unit of momentum is the bole. 
It is the momentum produced when a mass of 1 gram is 
moving at the rate of 1 cm. /sec. The number of boles is 
equal to the product of the mass in grams and the velocity in 
centimeters/sec. (b = Mv). For example, a 100 Kg. pro- 
jectile with a velocity of 300 m./sec. will have a momentum 
of 3,000,000,000 c.g.s. units or 3,000,000,000 boles (100 X 
1,000 X 300 X 100). 

49. Newton's Laws of Motion. — The following laws were 
formulated by Sir Isaac Newton. A thorough understanding 
of these laws is very important. 

/. Every body continues in its state of rest or of uniform motion 
in a straight line, unless compelled to change that state by an 
impressed force. 

II, Change of momentum is proportional to the force acting 
and takes places in the direction in which the force acts. 



MOTION 



39 



///. To every action there is always an equal and opposite 
reaction. 

50. Newton's First Law. — A body is incapable of putting 
itself in motion. A book, lying on a table, will remain in the 
same position indefinitely, unless acted upon by some outside 
force. Likewise, a body in motion will continue to move 
indefinitely, unless acted upon by some outside force. A 
thrown baseball tends to continue on forever in a straight 
line. Were it not for gravity and the retarding effect of the 
atmosphere, the ball would never come to rest. It is clear 
from this discussion that Newton's first law is a statement of 
the law of inertia, 

51. Newton's Second Law. — It is obvious from Newton's 
second law that the amount of change of momentum undergone 
by a body depends on the magnitude of the force acting. 
Thus we are able to measure a force by the change in momen- 
tum it will produce in a given time. The force varies directly 
as the change in momentum and 
inversely as the time consumed. 



H 




I 



F 



F aMv / t 

If the impressed force acts counter 
to the motion of the body, the 
momentum will be decreased; if 
with the motion of the body, the 
momentum will be increased. 

According to the second law, a 
cannon ball dropped from a height Fig. 29. — A and B will reach 
of 200 ft. will strike the ground at '^" ^^^' ^' '^^ '^^^" ^^^'^^*- 
exactly the same moment as another ball shot horizontally 
from the same height. This is due to the fact that each ball 
is given the same downward acceleration by gravity. Further, 
the balls will always keep the same relative position from the 
ground while in the air. We see also, from the illustration, 
that the change of momentum is always in the direction in 
which the force acts. The first ball is acted upon by gravity 
only and has simply a downward momentum. The second 



40 



MECHANICS 



ball is acted upon by the force of the explosion as well as 
gravity and has both a horizontal and downward momentum, 
each change of momentum being independent of the other. 

A simple experiment will illustrate the above law. A and 
B (Fig. 29, essential details shown onh') are two similar steel 
balls. B rests on the support F and A is held in place by the 
friction between C and G. The thumb screw D enables 
proper adjustment of the spring C attached at H. With the 
balls A and B at the same level, the hammer E is allow^ed to fall 
against C As a result A is released and falls to the floor, 
while B is driven out horizontally and describes a parabolic 
path to the floor. Each ball reaches the floor at the same 
instant. 

52. Newton's Third Law. — Any action between two bodies 
is mutual. In jumping from a boat to the shore, the boat 

tends to gain an equal and 
opposite momentum to that of 
the person jumping. A per- 
fectly elastic ball thrown 
against a brick wall tends to 
^~< \^^ bound back with the same 

velocity with which it struck 
the wall. In firing big guns it 
is necessary to absorb the reac- 
tion by means of recoil springs, 
otherwise the guns might break 
loose from their foundations. In case of an inelastic body 
like putty, the reaction will have a flattening effect. 

In Fig. 30, A and B are two highly elastic steel balls. If 
B is displaced to the position of 5' and allowed to fall against 
A J the impact will cause A to undergo an equal displacement 
in the opposite direction as shown in A\ In case A is held 
fast, B will bound back to B\ 

53. Graphical Representation of Motion. — A single force, 
if of sufficient magnitude, will cause a body to move in a 
straight line. The motion is fully defined when the following 
are given: (a) the amount of the motion; (6) the direction of the 



Fig. 30.- 



? .;=r=^^--^ 


---^ 


/ 




\ 


/ 




\ 


/ 




\ 


/ 




\ 


/ 




\ 




A B 

Action and reaction. 



MOTION 



41 




motion; (c) the former position of the body. To represent the 
motion of a body graphically, it is only necessary to draw a 
straight line of suitable length, with an arrow head pointing 
in the proper direction. Such a line is called a vector or 
vector line. To illustrate, suppose 
we wish to represent the motion 
of a body travelling east with a 
velocity of 10 ft. /sec. First we 
adopt some convenient scale, such 
as 1 in. = 2 ft. /sec. Then we 
draw a line 5 in. long, using an 
east and west axis. At the east 

extremity we draw an arrow head Fig. 3 1.— Velocities represented 

pointing due east. In construct- ^^^^ ^^^ ^' 

ing vector lines, the scale used should be indicated clearly. 

In Fig. 31, there are four vector lines indicating velocities 
from a common point in four different directions. Since the 
scale used is | in. = 10 ft. /sec, it is evident, by a measure- 
ment of the lines, that A has a velocity of 10 ft./sec. due 
west; B a velocity of 20 ft./sec. due southeast; C a velocity 
of 15 ft./sec. due east; and D a velocity of 5 ft./sec. due 
north. 

54. Velocity of Rotation. — In expressing the velocity of 
a rotating body two methods are in use : (a) the length of arc 
described by a selected point in a given time; or (6) the angular 

change that the point makes from the axis 
of rotation in a given time. The point 
selected is considered to be on the rim of 
the rotating body. In the first case, it is 
clear that the arc described will be in pro- 
portion to the distance to the center of 
rotation; while, in the second case, the 
Angle d is angular velocity will be independent of the 
distance from the center. 
Angular velocity may be expressed as revolutions per minute 
(r.p.m,) or degrees per minute (d.p,m.). In the latter case a 
unit called a radian is used. A radian is an angle of such 




Fig. 32 

a radian 



42 MECHANICS 

magiiitude that the arc which it intercepts on the circumference 
is exactly equal to the radius of the circle. 

In Fig. 32, angle ^ is a radian since the arc AB is just equal 
in length to the radius r. As the circumference of a circle 
equals 27rr, it will be seen that in every circumference there 
are 27r radians (27rr/r). Also, since there are 360° in a circle 
it is evident that a radian is equal to 57.3° (360/27r). Further 
the linear velocity of any point may he determined by multiplying 
the velocity in radians by the distance froni that point to the 
axis of rotation. If the angular velocity in Fig. 32 is 10 
radians/sec, the linear velocity of any point in the circum- 
ference will be 30 ft./sec. (10 X 3). 

Questions and Problems 

1. Define (a) motion; (6) rest. How is a body set in motion? 
Brought to rest? 

2. What is (a) absolute motion? (6) relative motion? Discuss fully. 

3. State the difference between speed and velocity and give an 
example of each. 

4. What is variable velocity? Average velocity? Instantaneous 
velocity? How would you compute the actual distance passed over in a 
pjiven time by a lx)dy whose velocity is variable? Uniformly accelerated? 

6. Define and give an example of (a) translatory motion; ih) rotary 
motion. Give an instance of a combination of the two. 

6. What is acceleration? Uniform acceleration? Varial)le accelera- 
tion? Positive acceleration? Negative acceleration? Illustrate in 
each case. 

7. What is the mathematical relation between velocity, uniform 
acceleration and the time? Between total distance passed over, average 
velocity and the time? Between total distance passed over, uniform 
velocity and the time? 

8. Define momentum. What is the formula for momentum? 

9. What is the unit of momentum in the f.p.s. system? The c.g.s. 
system? 

10. State and discuss Newton's first law of motion. 

11. State and discuss Newton's second law of motion. What general 
relation is there between momentum, the force producing it and the time? 
Describe an experiment to illustrate the second law. 

12. Stat« and discuss Newton's third law of motion. Give an experi- 
ment to show that action and reaction are equal and opposite in direction. 

13. What is a vector or vector line? How may the velocity of a body 
be represented graphically? Give an example. 



MOTION 43 

14. In how many different ways may the velocity of a rotating body 
be expressed? What is a radian? How many radians in a circum- 
ference? How many degrees in a radian? How may the velocity in 
radians be reduced to linear velocity? 

16. Represent graphically a velocity of 100 ft. /sec. acting north. 

16. Represent graphically velocities of 100 ft./sec, 80 ft. /sec, 70 ft./ 
sec. and 65 ft./sec. acting north, east, south and southwest respectively 
from the same point. 

17. An automobile flywheel is making 1,200 r.p.m. Express the 
velocity in degrees/min.; in radians/min. If the wheel is \}/^ ft. in 
diameter, what is the linear velocity of a point on the circumference? 

18. A locomotive is turned around (front to back) on a turntable in 
4 minutes. Express the angular velocity in radians/sec. In degrees/ 
hour. 

19. Two gyroscopes are running with velocities of 2,000 r.p.m. and 
200 radians/sec. respectively. Which has the greater velocity? By 
how much? Give answer in radians. 

20. A circular saw is running at the rate of 12,000 radians/sec. The 
distance between the extremities of two diametrically opposed teeth 
is 18 in. How far does the point of a tooth travel per second? How 
many r.p.m. does the saw make? 

21. Express the velocity of a point on the equator in r.p.m. In 
ft./sec. In radians/sec. Assume circumference of the earth to be 
25,000 mi. 

22. A grinder is running at the rate of 3,500 r.p.m. By how much does 
this vary from a speed of 200 radians/sec? Give answer in r.p.m. 



CHAPTER VII 
COMPOSITION OF FORCES AND VELOCITIES 

55. Resultant of Two or More Forces. — Whenever two or 
more forces in the same plane are appHed to a body at a 
fixed point, these forces may be replaced by a single force 
which will have the same effect on the body as the joint action 
of the original forces. This single force is known as the 
resultant force. The process of determining the resultant of 
two or more concurrent forces is known as the composition 
of forces. The separate forces are called components. 

For the sake of simplicity, let us assume that we are dealing 
with two forces only. It is evident that the two forces may 
act (1) in the same straight line in the same direction; (2) 
in the same straight line in opposite directions; (3) or at an 
angle with each other. 

1. If two forces act in the same straight line and in the 
same direction, it is evident that the resultant is equal to 
their sum. For example, if two forces of 10 and 20 lb. each 
are applied to a body and act due east, they may be replaced 
by a single or resultant force of SO lb. applied at the same point 
and acting due east. 

2. If two forces act in the same straight line and in opposite 
directions, it is evident that the resultant is equal to their 
difference. For example, if two forces of 10 and 20 lb. act 
east and west respectively, they may be replaced by a single 
or resultant force of 10 lb. applied at the same point and acting 
due west. 

3. If two forces act upon a body at an angle, the resultant 
is equal neither to their sum or difference, but must be found 
by means of the parallelogram law. This will be described 
in the following paragraph. 

44 



COMPOSITION OF FORCES AND VELOCITIES 



45 



56. Parallelogram of Forces. — Suppose it is desired to find 
the resultant of two forces acting at an angle and represented 
graphically as in Fig. 33. Using the force lines Fi and F2 
as adjacent sides, a parallelogram is constructed and a diagonal 




Fig. 33. 



-The resultant R has the same effect as the joint action of F] 
and F2. 



from the point of application is drawn. The diagonal repre- 
sents the resultant both in amount and direction, according 
to the scale used. Thus the single force R will have the same 
effect as the joint action of Fi and F2. 

57. Experimental Verification of the Parallelogram Law. — 
The following experiment was performed by the author as a 




Fig. 34. — Apparatus for verifying the parallelogram law. 

classroom demonstration, the purpose being to prove that the 
parallelogram law is true. A weight of 19.4 lb. was suspended 



46 MECHANICS 

before the blackboard as shown in Fig. 34, care being taken 
that there was no friction between the weight and the board. 
The balance readings recorded were 10 and 15 lb. We now 
have three concurrent forces in equilibrium. It is evident 
that the combined effect of the upward forces is just sufficient 
to hold up the suspended weight. In other words, the 
resultant of the two upward forces must be equal and opposite 
in direction to W, If this is confirmed by the parallelogram 
law, we may assume the law to be correct. 

The three forces were located on the blackboard by placing 
a rectangular block along the cords and marking with a sharp 
piece of chalk. The forces (10 and 15 lb.) were represented 
graphically, the scale used being 1 in. = 1 lb. The parallelo- 
gram was completed and the diagonal drawn. The diagonal 
measured 19.4 in. According to the scale assumed, 19.4 in. 
represents 19.4 lb. Since the resultant is found to be equal 
and opposite in direction to W, we may be certain that the 
parallelogram law is correct. W is here known as the equi- 
lihrant. The equilihrant is always equal in magnitude and 
opposite in direction to the resultant. 

58. Composition of Velocities. — Since velocities are brought 
about by the apphcation of forces, it logically follows that the 
resultant of two velocities may be determined in the same way 
as the resultant of two forces. It is thought that no discussion 
will be necessary to enable the student to solve such simple 
problems as will be given him. 

59. Forces and Velocities Exceeding Two in Number. — In 
case we have to deal with more than two forces acting at an 
angle, it is necessary to find the resultant of two of them; 
then to find the resultant of the resultant just found and the 
next force. This process is continued until each force has 
been used. The same method is followed in case of more than 
two velocities. 

60. Directions for Graphical Work. — Accurate work is 
essential in solving problems b}^ the graphical method. Care- 
lessness is inexcusable and results in a waste of time. The 
following directions and suggestions should be studied carefully. 



COMPOSITION OF FORCES AND VELOCITIES 47 

1. Graphical diagrams should be drawn in pencil. A fairlj^ hard pencil 
with a sharp point gives the best results. 

2. If the paper is of good grade, the lines may be inked over with 
india ink. Never use ordinary ink. 

3. Be sure that the ruler used has a straight edge. 

4. Use a compass in completing the parallelogram. 

5. Be sure that all angles are correctly represented. 

6. Avoid small parallelograms. It is suggested that one half of an 
ordinary sheet of paper be allowed. This will tend to increase accuracy. 

7. Use squared paper if possible. 

8. Label each diagram fully, being sure to indicate the scale used. 

Questions and Problems 

1. What is meant by the composition of forces? 

2. Define (a) resultant; (6) component; (c) equilibrant. 

3. How would you find the resultant of several forces acting at an angle 
and attached at the same point? 

4. Find the resultant of the forces given below: 
(a) 10 lb. and 30 lb.; angle between = 25°. 
(6) 100 lb. and 60 lb.; angle between = 40°. 
(c) 24 Kg. and 36 Kg.; angle between = 90°. 
{d) 100 lb. and 200 lb.; angle between = 125°. 
(e) 200 lb. and 112 lb.; angle between = 140°. 
(/) 500 g. and 590 g.; angle between = 100°. 

5. A motor boat has a velocity of 10 m.p.h upstream (north). The 
wind gives the boat a velocity of 2 m.p.h. to the east. Find the direction 
in which the boat goes and the actual distance covered in one hour. 

6. An automobile is travelling north at a rate of 40 m.p.h. The wind 
is blowing from the west at the rate of 15 m.p.h. What is the resultant 
velocity of the wind? From what direction does the wind seem to 
come? 

7. In what direction would a weather vane point in problem 6, if 
mounted on the radiator of the car? 

8. A balloon rises at the rate of 25 ft. per second. It is blown east at 
the rate of 5 ft. per second. Find the rate and direction of motion. 



CHAPTER VIII 



RESOLUTION OF FORCES AND VELOCITIES 



61. Resolution of Forces and Velocities. — It has been 
shown, in the previous chapter, that we may replace two or 
more forces (or velocities) applied at the same point and in 
the same plane, by a single force (or velocity) having the 
same effect. We shall now see that it is possible to replace 
a single force (or velocity) with two or more components 

having the same joint effect as the 
original force (or velocity) . The process 
of replacing a single force (or velocity) 
with two or more forces (or velocities) 
having the same joint effect as the original 
force (or velocity) ^ is called the resolution 
of forces (or velocities). Resolution is 
the opposite of composition. Problems 
dealing with resolution may be solved 
by means of the parallelogram law. 
Suppose we wish to resolve a force 
Fig. 35.— -P may be re- (F) into two components (Ci and C2) 

placed by its components ^^ i^^ ^f 3Q0 ^^^ 4Q0 respectively 

Ci and 62. -IIP T^' r, ' 1 

With the force. First F is represented 
graphically, both in amount and direction, as shown in Fig. 
35. Ci and C2 are drawn from of indefinite length and at 
the proper angles. The components are now represented in 
direction, but not in magnitude. A parallelogram is next 
formed with F as the diagonal. Thus the components (Ci and 
C2) are defined in length. According to the scale assumed 
for Fj they are now determined in amount. Ci and C2, acting 
jointly, will have exactly the same effect at as F. 

In case the components are known in magnitude, but not in 

48 




RESOLUTION OF FORCES AND VELOCITIES 



49 



direction, a pair of compasses will be necessary to complete the 
geometrical construction. 

If the angles are decreased, the components will decrease; 
if the angles are increased, the components will increase. 

62. Vertical and Horizontal Components. — Frequently it is 
necessary to resolve a given force or velocity into rectangular 
components. Figure 36 represents a force (F) resolved 
graphically into vertical and horizontal components (V and 
H). The vertical component (V) represents the effective 




Fig. 36. — 'Rectangular components. 

value of F vertically and the horizontal component (H) repre- 
sents the effective value of F horizontally. Thus we are able 
to determine both the lifting effect and the horizontal pull 
produced by F. 

63. Value of Squared Paper. — Graphical results in the 
resolution of forces and velocities are facilitated by the use of 



7?' 



Scale ^^'ij^ 



120^ 



-y' 



'A 



Z. 






^. 



~-Z: 



¥^se' 



3BSpace5 
Fig. 37. — Resolution by means of squared paper. 

squared paper. For example, suppose (Fig. 37) we have a 
force of 120 lb. acting </> degrees to the horizontal. If the 



50 MECHANICS 

paper is ruled in 3 20 ii^- squares, then 3^^o in- = 3 lb. is a good 
working scale. Thus to represent 120 lb., we draw a line 2 in. 
long. The rectangular components may now be determined in 
value directly, a rule being unnecessary. By inspection we 
see that V is 24 spaces long and = 24 X 3 = 72 lb. in magni- 
tude. H is found to be 32 spaces long and = 32 X 3 = 96 lb. 
in magnitude. 

64. Resolution by the Trigonometrical Method. — A force 

or velocity may readily be resolved into vertical and horizontal 

y f components by the use of simple 

^^^\ trigonometric functions. Refer- 

^^ \ ring to Fig. 38, it is evident that 

eO^^ I cos 30° = H/F, oiH=F cos 

^^ i 30°, // = 60 X .866 = 51.96 lb. 

'^y^ 30'' i Also cos 60° = F/F, ox V = F 

^ ^jg^4 ^^ COS 60°, F = 60 X .500 = 30 lb. 

liG. 38. — Rectangular components 65. CompOSitiOH of ForCeS by 

determined trigonometricaiiy. Resolution into Vertical and 
Horizontal Components. — If several forces act in the same 
plane and are apphed at the same point, their resultant may 
be found by resolving all the forces not vertical or horizontal 
into their vertical and horizontal components. Since the 
forces are now all vertical or horizontal, by proper additions 
and subtractions we may combine them into one vertical and 
one horizontal force. The resultant of the last-mentioned 
forces will given the resultant of the original forces both in 
magnitude and direction. 

Questions and Problems 

(Solve problems graphically unless otherwise instructed) 

1. Explain what is meant V^y resolution of forces. 

2. What are rectangular comix)nents? 

3. How would you find the resultant of several forces applied at the 
same point and in the same plane? 

4. Resolve a force of 20 Kg. into two components at angles of 15° and 
25° with the force. 

5. Resolve a force of 200 lb. into two components such that one com- 
ponent will be a force of 120 lb. acting at an angle of 30° with the 200 
lb. force. Give also the angle at which the second component acts. 



RESOLUTION OF FORCES AND VELOCITIES 



51 



6. Resolve a force of 40 lb. into two components of 20 lb. and 30 
lb. each and find the angle between each component and the original force. 

7. Resolve a force of 500 lb. acting at an angle of 35° to the horizontal 
into its vertical and horizontal components. Solve with squared paper. 

8. Repeat Problem 7, using a trigonometrical solution. 

9. Repeat as in Problems 7 and 8 for a force of 1,000 lb. acting at 
an angle of 60° with the horizontal. 

10. A motor boat has a velocity of 15 m.p.h. due north in still water. 
In one hour, due to a west wind, it actually travels 18 miles due northeast. 
Find the velocity imparted to the boat by the wind. 

11. A tug boat is towing a scow at an angle of 5° with the axis of the 
scow. If the tension in the tow line is 300 lb., what force is moving the 
scow ahead and what force is tending to move the scow to the side? 



6-^ 



4 



^IC'' 



6-^ 



(1) 



5' 






(2) 
Fig. 39. 




12. Using the method of resolution into rectangular components, 
determine the resultant of the sets of forces in Fig. 39. 

13. Resolve a force of 300 Kg. into two components acting at an angle 
of 80° with each other, such that one force shall be three times the other. 

14. What force will be necessary to keep a 100 lb. ball on an inclined 
plane 10 ft. long and 3 ft. high? Assume the supporting force to act 
parallel to the plane and at the center of the ball. 



CHAPTER IX 



^/^ 



B 



^H. 



Fig. 40.- 



Concurrent forces in 
equilibrium. 



EQUILIBRIUM OF CONCURRENT FORCES 

66. Concurrent Forces in Equilibrium. — Forces applied at 
the same point, or passing through the same point if extended, 

are called concurrent forces. 
' Concurrent forces will be in 

equilibrium (of rest or 
motion) when the opposite 
vertical forces are balanced 
and when the opposite hori- 
zontal forces are balanced. 
Stated differently, a body 
acted upon by concurrent 
forces will be in equilib- 
rium, when the resultant of 
all the forces acting is zero. 
For example, the body B (Fig. 40) is at rest. It is evident, 
then, that: 

7i = 72 and Hi = H2 

A body acted upon by a single force or by unbalanced 
forces can never be in equihbrium. In case the forces are 
not vertical or horizontal, the body acted upon will be in 
equilibrium if the resultant of the vertical components is 
zero and the resultant of the horizontal components is zero. 

Another case of equihbrium is illustrated in Fig. 41. A 
body (B) weighing 125 lb. is being moved at a uniform velocity 
along the surface S-Sl by a force of 50 lb. acting at 
an angle of 25° to the horizontal. Suppose we wish to find 
the force of friction (Fr.) and the perpendicular pressure 
(P) between the body and the surface. Referring to the 
'^free body'' diagram (Fig. 42), it is evident, since the 
body is in equihbrium, that: 

52 



EQUILIBRIUM OF CONCURRENT FORCES 



53 



(1) H = Fr. and (2) V + P = G. 
Since H = 50 cos 25°, and F = 50 cos 65°, it follows that 
(1) 50 cos 25° = Fr., and Fr. = 50 X .906 = 45.3 lb., 




* fk< 




Fig. 41. 



Fig. 42. — Force diagram for Fig. 41. 

(2) 50 cos 65° + P = 125, and P = 125 - (50 X .423) = 
103.85 lb. 
67. Triangle of Forces. — If three concurrent forces in the 




Fig. 43. 





same plane are in equilibrium, they may be represented in 
amount and direction by the sides of a triangle. 

Suppose Pi, P2 and W (Fig. 43) to be in equihbrium. Then 



54 



MECHANICS 



as shown in Fig. 44, W is equal to /?, the resultant of Fi and 
7^2. Still referring to Fig. 44, it is evident that Fi, F2 and 
W are proportional respectively to the sides of the triangle 
ABD, since: (1) Fi is proportional to BA; (2) 7^2 is propor- 
tional to BC and therefore to AD; (3) W = R and is therefore 
proportional to BD, but opposite in direction. Hence the 
original forces, Fi, F2 and W are represented both in amount 
and direction by the sides of the triangle BA^ AD and JDJS, as 
shown in Fig. 45. 

68. Concurrent Forces Applied at Different Points. — Con- 
current forces may be applied to a body at the same point 
or they may be applied at different points. In the latter 
case, the force hues will meet at a common point if extended. 





Fig. 46. — Weight of bar A-B is equal 
to the resultant of Fi and F2. 



Fig. 47. — Force diagram 
for Fig. 40. 



For example, AB (Fig. 46) is a rigid, uniform bar weighing 
W lb. and supported by cords and balances as shown. The 
weight (W) of the bar acts vertically downward at the center 
of gravity (C.G,), W is balanced by the tensions (Fi and F2) 
of the supporting cords. It is clear, since the bar is in equilib- 
rium, that W is numerically the same as the resultant of Fi 
and F2. The plumb line (P), suspended at C, determines the 
direction of W and will always pass through the center of 
gravit^^ If Fi and F2 are known, W may be determined; 



EQUILIBRIUM OF CONCURRENT FORCES 



55 



if W is known, F] and F2 may be determined. The force 
diagram for Fig. 46 is shown in Fig. 47. Fi and F2 (balance 
readings) are drawn to scale at the proper angle and a paral- 
lelogram formed. The resultant {R) is found, using the same 
scale, and will check with the weight of the bar. 

In case the supporting cords diverge, as in Fig. 48, the 
solution is similar to the one just described. W is numerically 




Fig. 48. — Weight of bar A-B is equal to the resultant of Fi and F2. 



the same as the resultant {R) of Fi and F2. The dotted 
parallelogram lines will always intersect along the plumb 
line (P). Figure 48 is a combined picture and force diagram. 
In case W is known and it is desired to find Fi and F^j the 
force diagram will be the same, the only difference being that 
W is drawn to scale, the parallelogram completed, and Fi 
and F2 computed. The computed values should check with 
the balance readings. 

Questions and Problems 

1. What are concurrent forces? 

2. Under what conditions will concurrent forces produce equilibrium? 



56 MECHANICS 

3. What is meant by the triangle of forces? Explain fully, using 
diagram. 

4. Describe a typical experiment involving concurrent forces applied 
at separate points. 

5. Find the amount and direction of a force that will produce equilib- 
rium with two forces of 30 and 50 g. acting at an angle of 40° with each 
other. 

6. A block weighing 5 lb. is being pulled uniformly along a surface 
by a horizontal force of 1 lb. Show by diagram the actions and reactions. 

7. A body weighing 50 lb. is being pulled uniformly by a force of 
12 lb. acting at an angle of 15° to the horizontal (See Fig. 41). Find 
the perpendicular reaction of the surface and the horizontal pull. 

8. A picture weighing 10 Kg. is suspended by two cords of equal 
length acting at an angle of 50° with each other. Find tension in the 
cords. 

9. A wire AB 30 ft. long is fastened at either end. If a man weighing 
160 lb. stands in the middle and the wire is depressed 2 ft., what is 
the tension at each end? 

10. A bar weighing 10 lb. is suspended as in Fig. 46. Find Fi and ft. 



CHAPTER X 

EQUILIBRIUM OF PARALLEL FORCES 

69. Tendencies of Parallel Forces. — Figure 49 represents 
a rigid, weightless body acted upon by parallel forces. Various 
elfects may be produced by the forces, according to their 
magnitudes and points of application. They may: (a) cause 
an upward displacement of the body AB; (b) cause a down- 
ward displacement oi AB; (c) cause AB to rotate; (d) cause a 




k--^/ --■>\<d,->H ^3 



->H^-> 



:5 



Fig. 49. — Body acted upon by parallel forces. 

simultaneous displacement and rotation. 

70. Equilibrium of Parallel Forces. — In order to avoid the 
effects as stated in the previous paragraph, that is, in order 
that the body AB shall be in equilibrium (at rest), the follow- 
ing conditions must prevail: 

1. The sum of the '^up'^ forces must equal the sum of the 
'^down^' forces, 

Fi + F\ must equal F2 + Fs 

2. The sum of the moments tending to produce rotation in 
one direction must equal the sum of the moments tending to 
produce rotation in the opposite direction. Assuming any 
point as the axis of rotation (fulcrum), e.g.^ point A^ then, 

F2 X (di + d2) + Fs X (di + d2 + da) must equal 

FiXdi+F,X (di + d2 + d, + d,) 
57 



58 



MECHANICS 



Summarizing, we may say that a body acted upon by 
parallel forces will be in equilibrium when : 

1. The Slim of the forces acting in one direction equals the 
sum of the forces acting in the opposite direction; 

2. The sum of the clockwise moments equals the sum of the 
counter-clockwise moments. 

71. The Laws of Parallel Forces Proven by Experiment. — 
The conditions necessary for equilibrium as stated in the 
previous paragraph may be verified in the laboratory. The 
following experiment was performed by two students. 



B C 



^G 



Wf.ofBar 
f 



IV, 



^i 



Fig. 50. — Apparatus for studying the laws of parallel forces. 



15* 



k': 



J. 



B C 



2M 5-x- /7 






: /3 



// 



6.?J' 



X--3 ---M 






Fig. 51. — Force diagram for Fig. 50. 



AG (Fig. 50) is a rigid steel bar weighing 6.25 lb. and sup- 
ported by two spring balances. Weights of 11.25 lb. and 10.75 
lb. were suspended from C and E. The weight of the bar con- 



EQUILIBRIUM OF PARALLEL FORCES 



59 



centrated at its center of gravity is represented by a force of 
6.25 lb. acting downward at D. The apparatus was carefully 
adjusted until AG was horizontal and the balances vertical. 
The balance readings were observed and the distances between 
the forces measured. A force diagram was then constructed, 
with the forces and distances clearly labelled (Fig. 51). 

The moments of the various forces were determined first 
about A and then about D, The results were finally tabulated 
as shown below. 



B 
C 
D 
E 
F 



Totals 



Forces in lb. 


Moments around A 
in Ib.-in. 


Moments around D 
in lb. -in. 


At 


Up 


Down 


Clockwise 


Counter- 
clockwise 


Clockwise 


Counter- 
clockwise 



15.00 


11.25 
6.25 


78.75 
150.00 


30.00 


330.00 





10.75 


397.75 




139.75 


13.25 






596.25 




28.25 


28.25 


626.50 


626.25 


469.75 



191.25 



278.25 



469 . 50 



An examination of the above tabulation shows that the 
sum of the ''up'' forces exactly equals the sum of the ''down'' 
forces; also that the sum of the clockwise moments is practic- 
ally the same as the sum of the counter-clockwise moments. 
The small difference in the moments is due to the fact that it 
is impossible to measure distances with 100 per cent, accuracy. 
This experiment verifies the laws of parallel forces as stated 
in Par. 70. 

72. The Resultant and Its Point of Application for a System of 
Parallel Forces. — Suppose it is desired to replace the system 
of forces in Fig. 52 by a single force (resultant) having the 
same effect as the joint action of the separate forces. We must 
find the magnitude of the force and its point of application, 
It is evident that the resultant will be equal and opposite 



60 



MECHANICS 



in direction to the force needed to cause equilibrium, 
force is found as follows : 

15 + ^ + 10 = 10 + 20 + 25 + 45, 
E = 75 lb. resultant force. 



This 



To find the point of application, assume A as the point of 

rotation and take distances 

from A as in the parentheses. 

From the law of equilibrium 



/5^ 



(3') 



10^ 

A 



(S'J 



12') 



10* 



(dJ 

I 
I 
I 

20* R 
Fig. 52. 



Le'j 



% 



25' 



45' 



of parallel forces we have : 

15 X 2 + 75 X d + 10 X 6 
= 20 X 3 + 25 X 5 + 45 
X7,d = 5.46 ft. from^. 

Hence we find that the re- 
sultant of the above forces is 



a down force of 75 lb. and that the point of application is 
5.46 ft. from^. 

73. Couple Defined. 



<- 



c/ 



Two equal and opposite 
forces acting at separate points constitute a couple. 
ring to Fig. 53, it will be seen that 
a couple causes rotation only and 
that there can be no resultant. In 
order to produce equihbrium, a second 
couple having an equal and opposite 
effect is necessary. The distance (d) 
between the forces is called the couple 
arm. The actual couple is Fi X d or 
^2 X d. 



parallel 
Refer- 



FiG. 53. — Couple. 



Questions and Problems 

1. State the conditions necessary for equilibrium of parallel forces. 

2. Describe a laboratory experiment to verify the above conditions. 

3. What is meant by the resultant of a system of parallel forces? 

4. What is a couple? Couple arm? What kind of motion does a 
couple produce? 

6. A painter's scaffold AB is 20 ft. long and weighs 100 lb. The 
painter, weighing 150 lb., is standing in the middle of the scaffold. Find 
the tension in each supporting rope. 



EQUILIBRIUM OF PARALLEL FORCES 61 

6. Repeat problem 5, if he painter is standing 5 ft. from A, 

7. Two men are carrying a load of 125 lb. on a uniform pole 20 ft. 
long. If the pole weighs 25 lb., where will the load have to be placed 
in order that one man shall carry twice as much as the other? Assume 
that the force exerted by each man acts 1 ft. from the end of the pole. 

8. A light and a heavy horse are harnessed together. Where must the 
king pin in the evener be placed in order that the light horse shall pull 
only % as much as the heavy horse? 

9. It is customary to use at least three horses on a reaper. Design 
a whiffletree for three horses, so that each horse shall exert the same 
pull. 

10. A uniform bridge 100 ft. long is supported by abutments A and B. 
The bridge weighs 175 tons and is carrying a locomotive weighing 60 
tons. If the center of gravity of the locomotive is 30 ft. from A, find 
the vertical reaction of each abutment. 

11. Repeat Problem 10, if the center of gravity of the locomotive is 
40 ft. from B. 

12. A uniform bar A B is 10 ft. long and weighs 20 lb. Three forces of 
8 lb. each act directly down on the bar which is horizontal. Two of 
the forces act at the extremity of the bar and the other 4 ft. from A. 
Find the resultant and point of application of the forces. 

13. Find the resultant and point of application, if the forces in Fig. 
52 are doubled and the distances remain the same. 

14. A door 7 ft. high and 3.5 ft. wide is supported by hinges 1 ft. 
from the upper and lower extremities. The door weighs 50 lb. Assum- 
ing each hinge to bear an equal load, what is the vertical and horizontal 
reaction at each hinge? 

15. The lever of the safety valve shown in Fig. 54 is 29 in. long. The 
lever weighs 3 lb. and its center of gravity is 14 in. from the fulcrum 



7/ — 1 



t 




Fig. 54. 



A. The valve is 2 in. in diameter. The valve stem is attached to the 
lever 3.5 in. from A. Weight of the valve and stem is 2 lb. A ball 
weighing 25 lb. is suspended from B. Make a free body diagram from the 
above figures and compute the force in lb. per sq. in. necessary to ^^blow^' 
the valve. 



CHAPTER XI 
EQUILIBRIUM OF NON-CONCURRENT FORCES 

74. Non-concurrent Forces in Equilibrium. — Forces which 
do not pass through a common point and which are not all 
parallel, are called non-concurrent forces. Since non-con- 
current forces may produce displacement as well as rotation, 
it is evident that such a system may not always be replaced 
by a single force or resultant. 

Non-concurrent forces will be in equilibrium when : 

1. The resultant of the vertical forces is zero, 

2. The resultarit of the horizontal forces is zero, 

3. The sum of the clockwise moments is equal to the sum of the 
counter-clockwise moments. 

If any of the three conditions stated above is not satisfied, 
there can be no equilibrium. The student should note that 
if the original forces are neither vertical nor horizontal, the 
forces should be resolved into their vertical and horizontal 
components. The components are then used as vertical and 
horizontal forces. 

75. The Ladder. — The ladder problem offers an excellent 
opportunity for the application of the laws of non-concurrent 
forces. AD (Fig. 55) is a uniform ladder 10 ft. long and 
weighing 14 lb. The rungs are spaced 1 ft. apart. The 
weight (W) of the ladder is considered as a force of 14 lb. 
acting downward at the center of gravity. A load (L) of 
40.25 lb. is suspended 3 ft. from D. The upper end of the 
ladder is just held free from the wall by a spring balance. 
Hence there is no friction between the ladder and the wall. 
The lower end of the ladder rests on a roller skate which in 
turn rests on platform scales. It is kept from slipping by 
means of a spring balance. The function of the roller skate 

62 



EQUILIBRIUM OF NON-CONCURRENT FORCES 



63 



is to eliminate friction between the ladder and the platform 
scales. The angle between the ladder and the wall is measured 
and found to be 44°. The spring balance readings are recorded 
also the reading of the platform scales (These values are used 
only for checking purposes). The balance readings are 17.5 
lb. each and the platform scales reading minus the weight of 
the roller skate is 54.25 lb. 




Platform Scales 
Fig. 55. — Ladder mounted for experimental purposes. 

It is now desired to find i/i, Hi^ F, the g;round reaction (G.R.) and 
the angle {6) of the ground reaction. It is evident that: 

1. Hi = H2. 

2. F = L + TF, 

3. W X BD sin 44° + L X CD sin 44° = H2 X AD cos 44°. 

As both Hi and H2 are unknown, it is necessary to take the moments 
about D as in (3). 

40.25 X 3 X .695 -f 14 X 5 X .695 = H2 X 10 X 719 
H2 = 17.52 1b. = Hi 
Since the observed readings Hi and H2 were 17.5 lb. each, it is clear 
that the computed and observed values check very closely. 
F = 40.25 4- 14 = 54.25 1b. 

The above value of F checks exactly with the observed value shown 
by the platform scales. 

The ground reaction (G.R.) is the resultant of F and Hi. G.R.- = 

flT' 4- F2, G.R. = VWi 2+ F2, G.R. = \/306.25 +2837.64 = 53.3 
lb. 



64 



MECHANICS 



Tan id) = V/Hi = 54.25/17.52 = 3.097. By referring to the 
trigonometric tables in the appendix, the angle in the tangent column 
corresponding to the decimal 3.097 is found to be 72°. 

76. The Wall Crane. — Figure 56 represents a laboratory 
model of a simple wall crane. The member AB (considered 
weightless) is held at rest by the tension {T) in the tie DBy 
the wall reactions (F and H) and the load (L). The com- 
pression in ^B is designated by the letter C Since AB is in 

equilibrium the following con- 
ditions prevail: 

1. H = C, 

2. H = T cos e, 

3. 7 + r sin 19 = L, 

4. r X 7 sin 19 = L X 4. 
Assuming L and to be given, 

T is found from equation (4). 
Substituting the value of T in 
equation (2), we get the values 
of H and C. Continuing the substitution in equation (3), 
the value of the vertical hinge reaction (F) is found. 

Questions and Problems 

1. What is meant by non-current forces? 

2. Under what conditions will non-concurrent forces be in equilibrium? 

3. A horizontal bar ^B 4 ft. long and weighing 10 lb., is hinged to the 
wall at A. B is supported by a tie attached to the wall at D, making 
an angle of 40° with the vertical. If 100 lb. is suspended 1 ft. from B^ 
find (a) the tension in the tie DB] (h) the compression in AB; (c) the 
vertical and horizontal reactions at A and D. 




Fig. 56. — Wall crane. 




T 4 , Y 



Fig. 57. 

4. In Fig. 57, find the tension in DB; the compression in AB; and the 
vertical and horizontal reactions at A and D. 



EQUILIBRIUM OF NON-CONCURRENT FORCES 65 

5. A uniform door 8 ft. high and 3.5 ft. wide, weighing 50 lb., is sup- 
ported by two hinges 1 ft. from the top and bottom respectively. 
Assuming the total weight of the door to be borne by the lower hinge, 
find the vertical reaction at the lower hinge and the horizontal pull 
exerted by the upper hinge. 

6. A ladder 20 ft. long makes an angle of 40° with a smooth wall. 
If the ladder weighs 35 lb. and its center of gravity is 9 ft. from the foot 
of the ladder, find (a) the horizontal reaction of the wall; (6) the vertical 
reaction at the foot of the ladder; (c) the friction between the ladder and 
the ground; (d) the ground reaction; (e) the angle of the ground reaction. 

7. Repeat Problem 6, assuming a friction of 6 lb. at the top of the 
ladder. 

8. Repeat Problem 7, if a painter weighing 175 lb. is standing 12 ft. 
up the ladder. 



CHAPTER XII 

COMMERCIAL AND LABORATORY STRUCTURES 

77. Trusses. — A truss is a system of members (beams, bars, 
rods, etc.), joined together by pins or rivets, to form a rigid 
framework. While trusses may be of various shapes, modern 
engineering favors those of the triangular type, — consisting 
either of a single triangle or a collection of triangles. Triangu- 
lar trusses are preferred on account of the fact that a triangle 
is the only geometrical figure which can not change its shape 





Fig. 58. — Simple truss. 



bed 

Fig. 59. — Compound truss. 



without changing the length of its sides. Trusses are so 
designed that the members are subject principally to 
tension or compression. As a rule, extraneous loads act as the 
joints. The material is so proportioned that deformations 
are practically negligible. Steady loads, as the weight of the 
material, etc., are called dead loads; varying loads, due to 
wind, snow, etc., are called live loads. 

Figure 58 represents a simple truss. The applied load (L) 
produces tension in CB and compression in AB. 

Figure 59 represents a compound deck truss supported at 
A and E, with loads (L) applied as shown. Tension members 
are shown with light lines; compression members with heavy 

66 



COMMERCIAL AND LABORATORY STRUCTURES 67 

lines. AE and bd are known as the upper and lower chords 
respectively. AB, BC, CD, DEj be and cd are called panels. 
Ay Bj Cj D, E, b, c and d are joints or apexes. The vertical 
distance h is the height or depth of the truss. The members 
between the chords, Ab, Bb, Be, Cc, Dc^ Dd and Ed, are web 
members or braces and are distinguished as verticals and 
diagonals. Web members in compression are usually called 
struts; tension members are usually called ties or stays. 

78. Roof Trusses. — Roof trusses are rigid structures of 
metal or wood or a combination of the two, designed to support 
roofs. A few of the standard types will be touched upon 
below. 

The Fink truss is commonly used for spans which do not 



(a) Fink. {b) Pratt. 



(c) Howe. (d) Triangular. 

Fig. 60. — Types of roof trusses. 

exceed 100 ft. Figure 60(a) shows a 40 ft. span. The Fink 
truss is very economical due to the relative shortness of its 
struts. The Pratt truss (Fig. 60(6)) is very popular and is 
often used instead of the Fink truss. The Howe truss (Fig. 
60(c)) is frequently used, especially if wood is to enter into the 
construction. The rafters and diagonals are generally of 
wood, the verticals of steel and the lower chords either of 
wood or steel. The triangular truss (Fig. 60 (rf)) is often used 
for short spans. 

79. Bridge Trusses. — Bridge trusses are divided into two 
classes: railroad bridges and highway bridges. A few of the 
more common types will be discussed below. 

Figure 61(a) represents a riveted Pratt railroad truss used 



68 



MECHANICS 



for spans up to 160 ft. The Warren riveted truss (Fig. 61(6)) 
is used for spans from 120 to 160 ft. It is cheaper than the 
Pratt truss and just as satisfactory. 

The quadrangular Warren truss (Fig. 62(a)) is commonly 




(a) Pratt. (b) Warren. 

Fig. 61. — Types of railroad bridges. 





(a) Quadrangular Warren. (6) Baltimore. 

Fig. 62. — Types of highway bridges. 

used for spans from 80 to 170 ft. The Baltimore truss (Fig. 
162(6)) is used for long span bridges. 

80. Laboratory Study of a Roof Truss. — ABC is a labora- 
tory model of a simple roof truss (Fig. 63). The members 




Fig. ()o. -Roof truss mounted for laboratory use. 

AB and CB weigh 2.5 lb. each and are hinged at B, The 
suspended weight W produces compression in AB and CB. 
The compression is balanced by the vertical and horizontal 
reactions at A and C. Inspection of the apparatus shows that 



COMMERCIAL AND LABORATORY STRUCTURES 69 

the compression in ^JB is the same as the compression in CB 
and that the reactions at A are the same as the reactions at C. 
It is assumed that one half of the weight of each member acts 
vertically downward at its extremities. Thus the total force 
acting downward at 5 is TT + 2.5 lb. Balance 1 gives the 
total downward force at A and balance 2 the horizontal reac- 
tion at A, The balances are used only as checks. 

The following experiment was done in the mechanics labora- 



Scale:l"=/S'' 




28.5' 
Fig. 64. — Force diagram for Fig. 63. 



tory of the Wm. L. Dickinson High School. A weight of 26 
lb. was suspended from B, making the total downward force 
26 + 2.5 or 28.5 lb. The apparatus was adjusted until 
Bi was vertical and B^ horizontal. Bi read 15.75 lb. and 
^2 read 15 lb. Angles ABW and CBW were measured and 
found to be 46° each. 

In order to find the compression in AB and BC and the 
vertical and horizontal reactions at A and C, a force diagram 
was constructed as shown in Fig. 64. Using a scale of 1 in. 
= 15 lb., a vector or force line was drawn to represent the 
total downward force of 28.5 lb. This force was resolved by 
the^parallelogram method into two components at angles of 



70 MECHANICS 

46° with the force. Each component was found to be 20.6 
lb. Thus the compression in AB and BC is 20.6 lb. 

It is evident that the compression in the member AB is 
just equal and opposite in direction to the resultant of the 
vertical and horizontal reactions V and H Sit A. Accordingly, 
the compressive force (20.6 lb.) was resolved into its vertical 
and horizontal components by the parallelogram method. 
It should be noted that the same scale is used throughout. 

V was found to be 14.5 lb. Since one half of the weight of 
the member AB acts at ^, it is necessary to add 1.25 lb. to V 
before checking w^ith Bi. Then the computed value of V 
checks exactly with the actual value as shown by the spring 
balance. H was found to be 15 lb., checking exactly with B2. 
As previously stated, the reactions at C are the same as at A 
and the compression in BC is the same as in AB, The results 
of the experiment were tabulated as follows. 

Weight of member AB 2 . 50 lb. 

Weight of member BC 2 . 50 lb. 

Weight suspended (TF) 26.00 lb. 

Total downward force at B 28 . 50 lb. 

Reading of balance 1 15 . 75 lb. 

Reading of balance 2 15 .00 lb. 

Angle ABW 46°. 

Angle CBW 46°. 

Calculated vertical component o^ AB 14.50 lb. 

}/2 weight of AB (acting on A) 1 . 25 lb. 

Total calculated downward force at A 15.75 lb. 

(check) Vertical force recorded by balance 1 15.75 lb. 

Difference , 

Calculated horizontal component of AB 15.00 lb. 

(check) Horizontal force recorded by balance 2 15.00 lb. 

Difference 

81. Laboratory Study of a Simple Truss. — Figures 65 and 
67 represent a laboratory model of a simple truss or, as it is 
often called, a *\stick and tie.'^ BC is a light compression 
member whose weight may be disregarded, fitting into a slot 
at B. The tie ^C is a tension member. Balance 1 reads the 
tension (7") in AC and balance 2 reads the compression in 



COMMERICAL AND LABORATORY STRUCTURES 71 

BC. These values are used as a check. Balance 1 is read 
directly. Balance 2 is pulled until BC just begins to leave 
the slot at B, The balance reading at this exact instant 
measures the compression in BC. 

The following figures were obtained by two students in the 
Wm. L. Dickinson High School. 

Case I. BC Horizontal 

A weight of 10.75 lb. was suspended from C and the appara- 
tus adjusted until BC was level. The angle ACB was meas- 



c;-^ Balance I. 





^^alance2 



JO.TS'f^ 



TcosACY^W 
or 



W 



7"= 



IV 



co6^CY 
N. Tco6ACB=R 



Fig. 65. — Simple truss mounted for Fig. 66. — Force diagram for Fig. 65. 
laboratory use. 



ured and found to be 35°. The balance readings were taken 
and recorded. Balance 1 was 18.5 lb. and balance 2 was 
15.25 lb. Using C as a free body, T and R were computed by 
trigonometry. Referring to Fig. 66, it is clear, since point 
C is in equilibrium, that : 

1. T cos ACY = W or T = W/ cos ACY, 

2, R = T cos ACB. 

First T was found in equation (1) and then its value was 
substituted in equation (2) to find R. 

1. T = W/ cos 55°, T = 10.75/.574 = 18.70 lb., 

2. R = T cos 35°, R = 18.70 X .819, = 15.31 lb. 



72 



MECHANICS 



Tabulation 

Angle ACB 35^ 

.Ajigle ACY 55^ 

Weight suspended (W) 10.75 lb. 

T (calculated value) 18. 70 lb. 

T (observed value, balance 1 ) 18 . 50 lb. 

Difference 20 lb. 

R (calculated value) 15 . 31 lb. 

R (observed value, balance 2) 15 . 25 lb. 

Difference 06 lb. 



Case II. BC not Horizontal 

In this case, a suspended weight of 10 lb. was used and the 
apparatus adjusted until BC was at an angle as shown in Fig. 
67. The angles BCW and ACB were measured and found 



Balance /. 




^ Balance 2 




Ficj. G7. — Simple tniss mounted for 
laboratory use. 



Tcos/^CD=RcosBCD 
Tco5 ACYi-Fcos BC^-W 

Fig. 68. — Force diagram for 
Fig. 67. 



to be 60° and 59"^ respectively. The balance readings were 
recorded as a check. Balance 1 read 10.3 lb.; balance 2 read 
10.25 lb. 

Using C as a free body, T and R were computed by trigono- 
metry. Referring to the force diagram (Fig. 68), we see, 
since point C is in equilibrium, that: 

1. T cos ACD = R cos BCD, 

2. T cos ACY + R cos BCW = W. 

Since T and R are both unknown, it is necessary to solve 
for them by the method of simultaneous equations. Putting 
in all known values, we have: 



COMMERCIAL AND LABORATORY STRUCTURES 73 

1. T cos 2r = R cos 30^ T X .875 = RX .866, 

2. T cos 61° + R cos 60° = W, T X .485 + i? X .500 
= 10. 

Simplifying and removing decimals : 

1. 875 T = 866 R, 

2. 485 T + 500 R = 10,000. 

Since T = 866 R/875, then, in equation (2) : 

485 X 866 R/875 + 500 R = 10,000, 
R = 10.32 lb. 

Substituting the value of R in equation (1) : 

875 r = 866 X 10.32, 
T = 10.25 lb. 

Tabulation 

Angle BCW 60° 

Angle ACE 59° 

Angle BCD 30° 

Angle ACD 29° 

Weight suspended (W) 10 lb. 

T (calculated value) 10.25 lb. 

T (observed value, balance 1) 10 . 30 lb. 

Difference . 05 lb. 

R (calculated value) 10 . 32 lb. 

R (observed value, balance 2) 10 . 25 lb. 

Difference . 07 lb. 

82. Laboratory Study of a Simple Hoisting Crane. — Figure 
69 represents a laboratory model of a simple hoisting crane. 
The angles and suspended weight (W) may be varied as 
desired. Given W, the weight of the boom AB and the 
necessary angles, the tensions (Ti and T2) may be computed, 
as well as the compression (R) in the boom. The problem 
is similar to that of the simple truss just studied. Since AB 
is fairly heavy, its weight may not be neglected. One half 
the weight oi AB must be added to W for the total downward 
force (L). 

Referring to Fig. 70 (force diagram for Fig. 69), it is evident 
that T2 has the same tension as W (not L). Hence T2 = W. 



74 



MECHANICS 



It is also evident that: 

(a) Ti cos 35° + 7^2 cos 53° = R cos 65°, and 
.(6) T, cos 55° + R cos 25° = T^o cos 37° + L. 
Since T2 = TF, then T^i and R are the only unknowns. Ti 
and /2 are found by means of simultaneous equations as illus- 
trated in the simple truss, previously studied. Ti is checked 
by the balance C and R is checked by pulling out on AB with a 
spring balance. 




Fig. 69. — Lal)()rat()ry 
type of hoisting crane. 




Force diagram for Fig. 69. 



83. Laboratory Study of Shear Legs. — Figure 71 represents 
a laboratory model of a pair of shear legs such as is found 
around docks, etc. It will be seen that equilibrium is produced 
by the tension in OD and OS and by the compression in the 
legs OA and OC. It is desired to find OD, OS, OA, OC and 
the vertical and horizontal foot reactions at A and C. The 
apparatus is set up as shown and the tension in OD is checked 
by the balance Bi. The compression in OA is checked by 
pulling on B2 along the line OA until A just begins to move 
from its support. The compression in OC is checked simi- 



COMMERCIAL AND LABORATORY STRUCTURES 75 

larly. One half the weight of each leg must be added to W 
for the .total load (L) at 0. 

Since the forces are not all in the same plane, assume OA 
and OC to be replaced by a single force OR in the plane DOW, 
It is evident that the imaginary force OR is the resultant of 
OD and OS. Given the total downward force at (L) and the 




Fig. 71. — Laboratory type of shear legs. 



angles a and 6, OD and OR are found graphically by the 
parallelogram method or trigonometrically. 

Inspection of Fig. 71 shows that OR (just computed) is the 
resultant of OA and OC. Since these forces are in the same 
plane, OR is resolved either graphically or trigonometrically 
into its components OA and OC. Thus the compression in 
the legs is determined. 



76 MECHANICS 

Questions and Problems 

1. WTiat is meant by a truss? 

2. Why is a triangular truss favored by engineers? 

3. Draw a compound deck truss and label each part. 

4. What is a roof truss? Make a drawing to show four common tj'pes. 

5. What is a bridge truss? Into two what classes may they be divided? 
Show, by drawings, two common types of each. 

6. Describe a laboratory experiment to determine the static forces in 
(a) a simple roof truss; (h) a ''stick and tie;'' (c) a hoisting crane; {d) a 
pair of shear legs. 

7. Find the compression in BA and BC and the vertical and horizontal 
reactions at A and C in Fig. 63, if the angles remain the same and W 
equals 20 lb. 

8. Solve for T and R in Fig. 65, if the angles remain the same and W 
equals 12 lb. 



.CHAPTER XIII 
ELASTICITY 

84. Elasticity Explained. — Whenever a body is acted upon 
by an external force, a change in the shape of the body is 
produced. The applied force may act in three ways: (1) 
it may stretch the body; (2) it may compress the body; (3) 
it may shear the body. If the body has not been loaded too 
heavily, it will return to its original shape as soon as the 
displacing load has been removed. This is a manifestation 
of elasticity, a property which all matter possesses. Steel, 
glass, wood, rubber, etc. are elastic. 

The amount of elasticity is measured, not by the ease with 
which the distortion is effected, but by the difficulty with which 
the distortion is effected. Steel is exceedingly elastic, while 
rubber is only moderately so. A unit force, applied to a 
unit length of steel, will produce much less distortion than the 
same force, applied to a piece of rubber of like length and 
cross-section. Steel, therefore, is more elastic than rubber. 

85. Elasticity Defined. — Elasticity is a general property 
of matter in consequence of which all bodies having undergone a 
change in shape, tend to resume their original shape as soon as 
the displacing force has been removed, 

86. Hooke's Law. — // the applied forces are not too large, 
elastic deformations of all kinds are directly proportional to 
the forces producing them. This law was first enunciated and 
demonstrated by Robert Hooke,^ an Englishman, and bears 
his name. 

Let us analyze the above law in order to understand clearly 
what is meant. Suppose that we have a piece of copper wire 
suspended from a rigid support and that the free end bears 

1 Robert Hooke (1635-1703). English physicist. Made many 
inventions in physical and astronomical instruments. 

77 



78 



MECHANICS 



a scale pan of sufficient weight to take all the kinks out of the 
wire. Suppose, further, that a pointer moving over a gradu- 
ated scale is attached a slight distance above the scale pan. 
If a weight of 5 lb. is placed in the scale pan and the pointer 
moves down 2 graduations, then a weight of 10 lb. will cause 
it to move down 4 graduations, and a weight of 15 lb. will 

cause it to move down 6 



'Rigid Suppori- 




graduations. In short, if 
we double the pull, we 
double the stretch; if we 
triple the pull, we triple 
the stretch. In each case, 
the pointer will return to 
its first position as soon as 
the weight is removed. 

The apparatus shown in 
Fig. 72 is very convenient 
for studying Hookers law. 
A piece of german silver 
wire 104.5 in. long and 
.0005 sq. in. in cross-sec- 
tion is firmly clamped at 
the top to a rigid wall sup- 
port (A). The opposite 
end of the wire carries a 
scale pan in which the 
various loads (]F) are 
placed. The pointer (Z)) 
is attached to the wire at 
B and is pivoted at C It will be seen that as B moves 
down, D will move up. In order to magnify the elongations 
of the wire and make more accurate reading possible, DC 
is made 10 times as long as BC, Hence, the movement of 
the pointer along the scale is always 10 times as much as 
the actual elongation of the wire. 

The following experiment was performed by a student in 
the Wm. L. Dickinson High School with the apparatus and 



SloJ-fediScrew 



1 



Pomier 



6feel5ca/e 



Fig. 72. 



Scafe Fan 

^'■''aricl Weights 



Apparatus to verify llooke's 
law for tension. 



ELASTICITY 



79 



material just described. First a straightening load of 10 lb. 
was placed on the scale pan to take the kinks out of the wire. 
This weight is not counted in the calculations and is called 
the ^^zero^' load. The pointer (D) was adjusted so that it 
was somewhat below the center of the scale (E) and the scale 
was moved until the pointer just coincided with an even scale 
division. A weight of 5 lb. was added to the scale pan and 
left 2 min. On being removed, it was found that the pointer 
returned exactly to its original position, showing the apparatus 
to be properly adjusted. If the pointer had not returned to 
its original position, there was either a loose connection or a 
sagging of the wall support. 

Again the pointer reading at ^'zero'' load was taken and 
recorded. A load of 2 lb. was now placed in the scale pan 
and, after 2 min., the pointer reading was carefully taken. 
The loads were increased, 2 lb. at a time, until 8 lb. had been 
added, the pointer reading being recorded in each case. The 
results were tabulated as follows : 



Loads in lb. 


Pointer readings 
in in. 


Apparent elon- 
gations in in. 


Actual elon- 
gations in in. 





2.75 






2 


3.00 


0.25 


0.025 


4 


3.25 


0.50 


0.050 


6 


3.50 


0.75 


0.075 


8 


3.75 


1.00 


0.100 



A graph was made from the above data and is shown on 
page 80. Note that the load-elongation Hne is straight, denot- 
ing a direct proportion. 

87. The Elastic Limit. — If the german silver wire, used in 
the experiment just described, had been subjected to further 
load, a point would have been reached at which the loads and 
deflections would have not been in direct proportion. The 
stretches would then increase in greater proportion than the 
loads, and the load-elongation line would bend to the right. 



80 



MECHANICS 



The pointer would not return to its original position when the 
loads were removed, and eventually the wire would break. 

The stress beyond which an elastic material will not return to 
its original position when the load is removed, is called the elastic 
limit for that material. The elastic limit is difficult to deter- 
mine exactly, but may be determined very closely. 

88. The Yield Point. — After a material passes its elastic 
limit, a stress is reached at which the elongations continue 
without the addition of further load. This is called the yield 
point and shows that rupture is about to take place. 



to 

o 
a 



D 
O 



















1 






















_/ 


A 


















r/ 


/ 
















A 


ly 


/^ 
















yp 


4\ 


jT 
















i 


















r/ 


/\ 
















/ 


/^ 


\ 














/ 



















0.050 
S+retches in Inches 



0.100 



Fig. 7.3. — Graph showing relation of load and stretch for german silver wire. 

89. Stress. — Whenever an elastic body is acted upon by 
a force, it is said to be under stress. The stress ma}^ be the 
result of a tensile load, in which the molecules tend to separate; 
it may be the result of a compressive load, in which the mole- 
cules tend to crowd together; or it may be the result of a 
shearing load, in which certain particles are caused to slide 
past others. In every case, the body resists the effort to 
change its molecular arrangement. 

The internal resistance of a body to a change of shape is called 
stress. It is determined in amount by dividing the external 
force acting in lb. by the area in sq. in. over which the force 
acts. The area must always be taken at right angles to the 
force. Let us take a practical illustration. 



ELASTICITY 81 

Suppose a steel wire .10 sq. in. in cross-section is bearing a 
suspended load of 50 lb. The load of 50 lb. is evenly dis- 
tributed over the area of the wire. If we wish to use a wire 
of 1 sq. in. cross-section and keep the same rate of tension, we 
must use a load of 500 lb. In either case, the wire is under a 
stress of 500 lb. per sq. in. The stress is simply the load applied 
in lb. per sq. in.^ 

r,. external force actinq in lb. 

Stress = ^- ^ T.^^r^r—j' — 7— 

area m sq. m. over which the jorce acts 

90. Strain. — Whenever a body is subjected to a stress, a 
change of shape results. The amount of change of size per 
unit of original size is called strain. To illustrate : suppose we 
have a piece of copper wire about .04 in. in diameter and 72 
in. long. Under a load of 5 lb., the wire will elongate .019 in. 
If a load of 10 lb. is used, the wire will elongate .038 in. In 
the first case, the strain will be .019/72 or .000263 inch 
per inch.i In the second case, it will be .038/72 or .000526 
inch per inch.^ 

^ . _ change of size in inches 
original size in inches 

91. Young's Modulus. — The mathematical relation of the 
stress and the strain is called Young^s modulus. 

■^7- • -, 1 stress 

Younq s modulus = — — r- 

strain 

Young^s modulus is also called the coefficient or modidus of 
elasticity. It is practically the same for tension as compres- 
sion. The modulus for shearing strains will be considered 
later. The modulus for compression and tension, in the case 
of steel, will average about 30,000,000 lb. per sq. in.; for 
copper about 15,000,000 lb. per sq. in. These figures will 
vary somewhat, but maybe taken as very close approximations. 

We have seen, from the experiment with the german silver 
wire, that elastic deformations are directly proportional to the 
forces producing them, provided the elastic limit is not ex- 

1 May also be expressed in Kg./cm^. 

2 May also be expressed in cm. /cm. 



82 



MECHANICS 



ceeded. This was stated as Hookers law. We may now state 
Hooke's law thus: Within the elastic limit, the strain is directly 
proportional to the stress. Figure 74 shows this relation graph- 
ically. The data was taken from the experiment with the 
german silver wire and the stress and strain for each load was 
computed. The following graph should be carefully studied 
and thoroughly understood by the pupil. 



cr 

J- 
CD 
Cl- 

c 

O 

a. 



16000 








































y 


/ 




















y 


/ 


















f 


f 


/^ 








8000 










f\ 


















t' 




















V 


/ 




















/ 


X 


















n 


^ 























0.0005O O.OOIOO 

Inches per Inch 

Fig. 74. — Graph showing relation of stress and strain for german silver wire. 

92. Ultimate Strength.— The ultimate strength of an elastic 
material is the load in lb. per sq. in. necessary to break or rupture 
the material.^ The rupture may be due to a tensile stress, a 
compressive stress or a shearing stress. Suppose a steel bar 
.5 sq. in. in cross-section ruptures under a load of 50,000 lb. 
The ultimate strength is 50,000/.o or 100,000 lb. per sq. in. 
The ultimate strength for a body under tension is often called 
tensile strength. The tensile strength of mild steel will average 
about 70,000 lb. per sq. in. It will vary somewhat due to 
composition, temperature, etc. 

93. Factor of Safety. — The ratio of the breaking load to 
the worki?}g load for any material, is called the factor of safety 
for that material. The factor of safety will necessarily depend 
on the nature of the load. Under a steady load, steel should 
have a factor of 4; under a varying load a factor of 6; under a 

1 May also be expressed in Kg. /cm. 2. 



ELASTICITY 



83 



load in which the stresses are hable to occur sharply, a factor 
of 10. In machines subject to sudden overloads, the factor 
of safety is made very high to avoid any chance of fatal acci- 
dents. It is evident that all members which have to bear 
frequent and sudden overloads should have a high percentage 
of overstrength. 

94. Elastic Fatigue. — Practical experience and experiment 
teaches us that an elastic material is more liable to break 
under a repeated load than a steady load of the same size. 
This is especially true when the stresses consist both of ten- 
sions and compressions. The tendency to rupture under 
repeated strain is due to what is known as elastic fatigue and 
results in the destruction of the cohesive force at the point of 
rupture. The rear axle of an automobile occasionally breaks. 
This is generally due to repeated twisting strains. A piece 
of copper wire twisted back and forth will soon undergo 
elastic fatigue and break at the point of strain. 

95. Shear Explained. — When a body is subjected to a 



7^-^Z 




Fig. 75. — Diagram to illustrate simple shear. 



shearing stress, certain particles, of which the body is com- 
posed, tend to slide by other particles. This is well illus- 
trated in the case of a pair of metal shears, the metal being 
sheared apart where the blades come together. Planing, 
turning and punching are shearing actions. 

In order to understand the nature of a simple shear, let 
us refer to Fig. 75. Here we have a rectangular, elastic body 
ABC 4 X 4 in. on the top face and 2 in. high. Assume the 



84 MECHANICS 

face ABC to be attached rigidly to the surface on which it 
rests, so that there can be no movement at the bottom. A 
force of 80 lb. is attached to the face EFGH and is evenly 
distributed over the 16 sq. in. This force (L) will move the 
horizontal layers to the right, the movement varying from 
maximum at the top to zero at the bottom. Each succeeding 
upper layer will have a greater displacement to the right. 
Actually, there is also a bending tendency present. It is 
sufficient for our purpose, however, to consider only the simple 
shear, leaving the rest to more advanced books. 

96. Shearing Stress. — Shearing stress is equal to the force 
acting in lb. divided by the area in sq. in. over which the force 
acts. In Fig. 75, the shearing stress is 80/16 or 5 Ib./in.^ 
It may also be expressed in Kg./cm.^ 

97. Shearing Strain. — Shearing strain is equal to the 
horizontal movement in inches of a particle of a body 1 inch above 
the plane to which it is attached. Referring to Fig. 75, let 
AJ = 1 inch and JJ^ = .0006 inch. The shearing strain is 
therefore .0006 inch per inch. Suppose that EE' = .001 
inch and that AE = 3 inches. The shearing strain will then 
be .001/3 or .00033 inch per inch. Shearing strain may also 
be expressed in cm. ^cm. 

98. Modulus of Rigidity. — The modulus of rigidity (also 
called the modulus of elasticity for shear) is the ratio of the 
shearing stress and shearing strain. 

T,r 1 1 r ' ' T. shearing stress 

Modulus of rigidity = —, r^ — ,— .-* 

shearing strain 

99. Practical Illustrations of Shear. — The rivets used in 
fastening together boiler plates, etc., are subject to shearing 
stresses. Figure 76 shows a single-riveted lap joint. The 
rivet is subject to a shearing stress at ^'a^' and will fail at that 
point if the stress is of sufficient magnitude. This is an 
example of single shear. 

Figures 77 and 78 represent single and double-riveted 
butt joints. The rivets are liable to failure both at *'&^' and 



ELASTICITY 



85 



''c'' and are said to be in double shear. Figure 79 shows 
rivets having failed under single and double shear respectively. 

Illustrative ProUem.—Whsit force is required to punch a M in. hole in a 
steel plate M in. thick? Shearing strength of the material is 40,000 

lb./in.2 

Solution.— AreEi under shear = .75 X 3.1416 X .50 = 1.178 sq. in. 
1.178 X 40,000 = 47,120 lb. Arts. 

Illustrative Problem.— The head of a 1 in. machine bolt is ^i in. thick. 
Find (1) the force necessary to shear the bolt off at right angles to its 



<^ mm^ 



i 

a-' 
Fig. 76. — Single-riveted 
lap joint. 



w£\Ml 






^ 



m 



X^y 



^y 



^ 



b c 

Fig. 77. — Single-riveted 
butt joint. 




b' c 

Fig. 78. — Double-riveted butt joint. 



E 



□ 



Fig. 79. — Failure 
of rivets due to single 
and double shear. 



length; (2) the shearing stress tending to strip the head from the bolt 
due to an applied force of 10,000 lb. Shearing strength of the material 
is 40,000 lb./in.2 

Solution.— (1) Area under shear = 1 X 1 X .7854 = .7854 sq. in. 
.7854 X 40,000 = 31,416 lb. Ans. 

2. Area under shear = 1 X 3.1416 X .75 = 2.356 sq. in. 10,000/ 
2.356 = 4,244 Ib./in.^ Ans. 

Questions and Problems 

1. Define elasticity. 

2. What effects may be produced upon an elastic body by an external 

force? 

3. State and explain Hooke's law. 

4. Define stress, strain, elastic limit, yield point. 
6. What is meant by Young's modulus? 

6. Describe an experiment to determine the relation between stress 
and strain for an elastic body. 



86 MECHANICS 

7. Define ultimate strength, tensile strength, compressive strength, 
shearing strength, factor of safety, elastic fatigue. 

8. Define shear and give examples of shearing actions. 

9. Show, using diagram, how to determine stress and strain for simple 
shear. 

10. What is meant by the modulus of rigidity? 

11. A steel strut having a cross-section of 4 sq. in. is subjected to a 
compressive force of 2 tons. Compute the stress. 

12. A uniform steel bar 2 sq. in. in cross-section ruptures under an 
applied tensile force of 200,000 lb. What is the ultimate strength? 

13. If the above bar is of wrought iron, how large a force will be 
required for rupture? Tensile strength of wrought iron = 50,000 lb./ 
in. 2 

14. A wire 2 mm. in diameter breaks under a load of 100 Kg. If the 
diameter is increased to 4 mm., what load will be necessary for rupture? 

15. A steel rod 6 ft. long and 1 sq. in. in cross-section is subjected to a 
tensile load of 10,000 lb. Find the elongation in inches. Modulus of 
elasticity = 30,000,000 lb. /in. 2 

16. A hard, drawn copper rod 10 ft. long and 1 sq. in. in cross-section 
stretches .100 in. Find the tensile load necessary to produce the 
elongation. Modulus of elasticity 17,600,000 lb. /in. 2 

17. A steel bar 1 sq. in. in cross-section and 6 ft. long stretches .024 
in. under an applied load of 5 tons. Determine the modulus of elasticity 
for the material. 

18. A horizontal beam 4 sq. in. in cross-section is securely fastened at 
one end. If a load of 10,000 lb. at right angles to the beam is applied 
10 in. from the attached end, find the deflection of the beam at the 
point where the load is applied. Modulus of rigidity is 12,000,000 lb. /in. 2 

19. A steel plate Y^ iri- thick has an ultimate shearing strength of 40,000 
lb. /in. 2 What force is required to punch a 3^ in. hole through the plate? 

20. Repeat problem 19 for a % in. hole. 

21. A wrought iron rivet 3^^ in. in diameter is sheared off at right 
angles to its length. If the shearing strength of wrought iron is 40,000 
lb. /in. 2, what was the applied force? 

22. The head of a 1 in. steel machine bolt is % in. thick. A force of 
12,000 lb. is applied in such a way that there is a tendenc}' to strip the 
head from the bolt. Find the shearing stress due to the applied load. 

23. In problem 22, if the force is increased gradually, will the bolt 
fail due to a tensile stress or a shearing stress? Explain. 



CHAPTER XIV 
WORK 

100. Work Defined and Explained. — Work is the result 
accomplished when a force acts through a distance. The force 
may keep the motion of a body constant; increase, decrease or 
change the direction of the motion; or cause the body to 
undergo a change of shape or size. Common examples of 
work are a hoisting engine lifting concrete; a man climbing a 
mountain; a horse drawing a wagon; punching holes in sheet 
iron; stopping an automobile by means of the brakes, etc. 
Various other examples will occur to the student. The 
foundation of a building, although it exerts an enormous 
upward force, does no work, because the weight supported 
by the force is not moved. Force, then does not always result 
in work. This point should always be kept in mind. 

101. How Work is Measured.^ — The work done upon any 
body is measured by the product of the force acting and the 
distance through which the force acts. 

Work done = Force X Distance = F X S 

102. Units of Work. — Various units may be used to express 
the amount of work done. If a 10 lb. weight is lifted verti- 
cally 10 ft., 100 ft. -lb. of work results. The same work may 
be expressed as 1,200 in. -lb. or .05 of a ft.-ton. Further, a 
weight of 1 Kg. lifted vertically through a height of 10 meters 
results in 10 kilogram-meters of work. 10 Kg.-m. may also 
be expressed as 10,000 g.-m. or 1,000,000 g.-cm. The more 
common units are given below. 

UNITS OF WORK 

(a) The erg is the work done when a force of 1 dyne^ acts 

1 1 dyne = .00102 g. 1 lb. = 445,000 dynes. 

87 



88 



MECHANICS 



through a displacement of 1 centimeter. Since the erg is a 
very small unit, the joule is often used. A joule is equal to 
10,000,000 ergs. 

(b) The kilogram-meter is the work done when a force of 1 
kilogram acts through a displacement of 1 meter. 

(c) The foot-pound is the work done when a force of 1 pound 
acts through a displacement of 1 foot. 

103. Time Not a Factor in Work. — The element of time is 
not considered in an expression of work. In a scientific sense, 
work refers only to the result accomplished, regardless of the 
time consumed in obtaining the result. For example, it will 
require 20,000 ft.-lb. of work to raise 1,000 lb. of pig iron 
through a vertical height of 20 ft. It is evident that the 
amount of work will be the same if the iron is lifted all in one 
operation or if each pig is lifted separately. 

104. Further Discussion of Work. — If a body weighing 
100 lb. is raised vertically 10 ft., 1,000 ft.-lb. of work is 

done. If the same body 
is pulled 10 ft. horizontally 
across a floor, less work 
is done as the force re- 
quired to pull is less than 
the force required to lift. 
The pulling force will 
depend on the friction 
between the bodies in contact, and is always less than the 
lifting force. 

In case it is desired to com- 
pute the work done by a force 
acting at an oblique angle, it 
is necessary to find the com- 
ponent of the force in the 
direction in which the body 
moves. Suppose (Fig. 80) 
it is desired to find the work 

done by a force of 50 lb., acting at an angle of 30°. The 
body is moved horizontally 10 ft. The work done is 7iot 



Dispfacemeni- = /O 
> 




SO 



EFFECTIVE FORCE ^ 
SO cos 30 "^^ 43.3 Lb. 

Fig. 80. — Work done equals 433 ft.-lb. 




Fig. 81 



WORK 



89 



500 ft.-lb. It is 433 ft.-lb. and is computed as follows: 

The horizontal component of 50 lb. = 50 X cos 30° = 50 X .866 = 
43.3 lb. 43.3 X 10 = 433 ft.-lb. 

The work done in moving a body uniformly up a frictionless 
inclined plane may be determined in two ways. Suppose, 
referring to Fig. 81, that a force of 5 lb. is necessary to move a 
body weighing 10 lb. up a plane 20 ft. long and 10 ft. high. 

The work = weight X height = 10 X 10 = 100 ft.-lb. The work 
also = force X length = 5 X 20 = 100 ft.-lb. 

105. Work Diagrams. — It is often convenient to represent 



















^,- 


-X 














II 

1 






















































^ 







J space = Iff.'' 
Fig. 82. — Diagram representing 40 ft.-lb. of work. 

work done by a wor]z diagram, — in which the amount of work 
is proportional to the area of the diagram. Suppose we wish 
to represent a constant force of 8 lb. acting through a distance 
of 5 ft. Adopting some convenient scale (such as 1 space = 
2 lb. and 1 space = 1 ft.), we construct a diagram as shown 
in Fig. 82. It is evident that each square space = 2 ft.-lb. 
and that the total number of square spaces represent the entire 
work done or 40 ft.-lb. (20 X 2). 

Figure 83 represents the net work done by the head end of a 
Corliss steam engine cylinder on 1 sq. in. of the piston during 
one revolution of the crank shaft. The diagram was taken 
by means of a device called a steam indicator. The line 
abcdea represents the steam pressure at any point in the out- 



90 



MECHANICS 



ward and return stroke. It is evident that the effective 
pressure (P) at any chosen place will be proportional to a 
vertical Hne (ordinate) connecting the upper and lower 
extremities. By drawing a series of such ordinates at equal 
intervals and averaging them, the mean effective pressure for 
the entire working stroke can be determined, if the vertical 
scale is known. Since 1 in. = 70 lb., the mean effective 




Fig, 



Stroke ^2 f^ 

83. — Diagram representing work done by the crank end of a Corliss 
steam engine cylinder during one revolution of the crank. 



pressure will be about 35 Ib./sq. in. for the diagram shown 
here. The stroke is 2 ft.; hence the useful work on one sq. in. 
of the head end of the piston for one revolution will be 70 ft.- 
Ib. (35 X 2). 70 ft. -lb. also represents the area (average 
height X length) of the diagram in terms of the scale used. 

The student should note that : a = steam admitted into the 
cylinder;?) = steam is cut off; c = exhaust opens; d = exhaust 
stroke begins; and e = exhaust closes and compression begins. 



Questions and Problems 

1. What is meant by work? Give various examples of work 

2. What is the general formula for work done? 

3. Name and define the units of work. 

4. Discuss *'time as an element in work." 

6. State two ways of computing the work done on an inclined plane. 

6. A force of 1 dyne acts through 10 cm. How much work is done? 

7. An automobile weighing 2,000 Kg. travels 1 Km. up a 2 per cent, 
grade. Find the work done in Kg.-meters. 



WORK 91 

8. An elevator weighs 1,500 lb. more than its counter-weight. How 
much work is done if it rises 50 ft. ? 

9. A block of wood weighing 4,200 lb. is 8 ft. long, 4 ft. wide and 3 
ft. thick. If the block is lying on the 8 ft. X 4 ft. side, how much work 
is necessary to turn the block on the 3 X 4 ft. side? 

10. How many foot-pounds of work will be done by a gasoline engine 
of 15 per cent, efficiency on 10 gallons of gas? Assume 2,000 British 
thermal units per lb. of gas (1 B.t.u. = 780 foot-pounds). 

11. A hoisting engine burns 1 ton of coal during the day. If the engine 
is 2 per cent, efficient, how much work will it do in a day? (1 lb. coal = 
13,000 B.t.u.) 

12. A force acts through 10 ft. against a varying resistance. The 
successive resistances in pounds at the beginning of each foot are as 
follows: 50, 55, 60, 58, 51, 52, 63, 60, 57, 49, 51. Construct a work 
diagram, drawn to scale, and determine the amount of work done. 



CHAPTER XV 
POWER 

106. Power Defined. — We have seen that time is not an 
element in work. For example, it is possible, under certain 
conditions, for a light horse to do as much work as a heavy 
horse, provided the light horse is allowed more time to com- 
plete his task. The heavy horse is said to possess the greater 
power^ however, as he can do the work in a shorter time. 
Power concerns itself both with the work done and the time, 
A locomotive will do many times as much work in a given 
time as a hoisting engine and therefore has the greater power. 
Power is an expression of the rate at which work is done. 

^ ivork done 

rower = — — 

time 

107. Units of Power. — The units of power in general use 
are the horsepower^ the watt and the kilowatt. The watt and 
kilowatt are coming into popularity very rapidly. Electric 
light bulbs are rated entirely in watts. Electric motors and 
marine engines are commonl}^ rated in kilowatts. The horse- 
power is still widely used and is very important. 

108. The Horsepower. — James Watt^ estimated that a 
heavy work horse, traveling at an average rate of 2.5 miles per 
hour, could lift a weight of 150 lb. by means of a rope and 
pulley. This is equivalent to 33,000 ft. -lb. of work per minute 
or 550 ft. -lb. per second. Watt's estimate is probably inaccu- 
rate, yet it furnishes a very satisfactory basis for comparisons 
of power. Whenever work is being done at the rate of 33,000 
ft.-lb. per minutey one horsepower is being expended. 

^, ftAb. of work per minute 
Horsepowers 33-^^^ 

1 James Watt (1736-1819). Scottish engineer. Invented the first 
condensing steam engine and the first centrifugal governor. First used 
the steam indicator to determine the amount of work done by a steam 
engine. 

92 



POWER 



93 



109. The Watt and Kilowatt. — The power unit in the c.g.s. 
system is an erg per second. For practical purposes, it is 
customary to use a joule per second (10,000,000 ergs per 
second). The latter unit is called a watt in honor of James 
Watt. One horsepower is equivalent to 746 watts. Since 
there are 1,000 watts in a kilowatt, it is evident that one 
horsepower is nearly equivalent to ^i of a kilowatt and, con- 
versely, that one kilowatt is slightly more than ^^ of a horse- 
power. Electrically, the watt is the work done in one second 
by a current of one ampere flowing under a pressure of one 
volt. 

1 horsepower (hp.) = 746 watts 
1 kilowatt (kw.) = 1,000 watts 
1 horsepower = % kw. (approx.) 

1 kilowatt = IJ^^hp. (approx.) 




Fig. 84. — Prony brake mounted for power test of an electric motor. 

110. Brake Horsepower (b.hp.). — The brake method is 
used in determining the power deHvered by a rotating shaft or 
pulley. It is also known as the absorption method, since the 
power transmitted is absorbed and the mechanical energy 
of the rotating body is transformed into heat energy. A device 
called the prony brake is used to determine brake horsepower. 
It is described in the succeeding paragraphs. 

The following experiment was performed by a student in the 
Dickinson High School and serves to illustrate the construc- 
tion and operation of the prony brake. Referring to Fig. 
84, M is an electric motor rated to give 2 hp. at 1,500 r.p.m. 
and 108 volts. In order to check up the rating, the motor is 



94 



MECHANICS 



fitted with a hollow iron pulley (P), which is rotated against 
the friction of the adjustable brake (B). The brake is lined 
with some heat resisting material, such as is used for brake 
lining of an automobile. It carries an arm (A) which rests 
on a knife edge (K), The knife edge rests on the platform 
balance (J). The power is turned on and the thumb screws 
{S and >S') are tightened until the speed of the motor is approxi- 
mately 1,500 r.p.m. in a clockwise direction. Meanwhile, a 
stream of water is directed against the pulley to keep it from 
overheating. From the following final readings, the horse- 
power was computed. 

Force (F) exerted by arm on balance 7 lb. 

Length (L) of arm 1 ft. 

Revolutions per minute (r.p.m.) 1,530 



B.hp. = 



F X 2L X 3. 1416 X r.p.m. 
33,000 



^ , 7X2 X 3.1416 X 1,530 ^ ^. , 
B.hp. = 33^^^^ = 2.04 hp. 

It will be noted that the numerator of the above fraction 
is the foot-pounds of work which it is 
assumed the force (F) would do each 
minute, in rotating through a distance 
of 2L X 3.1416 every revolution for 1,530 
revolutions. 

Another form of the prony brake is 
shown in Fig. 85. B represents the 
balance wheel of an automobile from 
which the clutch has been removed. A 
belt of heat resisting material is carried 
around the wheel and secured at either 
f end by two spring balances (Ti and T2)* 
The throttle is opened and the balances 
tightened until the wheel is making 1,000 r.p.m. The balance 
readings are taken and the difference in tension noted. The 
wheel is kept from overheating as in the previous experiment. 
From the following data the horsepower is computed. 




Fig. 85.— Belt form 
prony brake. 



POWER 95 

Diameter (d) of wheel 1 . 5 f t. 

Difference in tension (T2 - Ti) 200 lb. 

Revolutions per min. (r.p.m.) 1,000 

Width of belt 25 in. 

^ {T2 - Ti) X d X 3. 1416 X r.p.m. , 
^^' 33,000 
^, 200 X 1.521 X 3.1416 X 1,000 ^^ , 
^'^^' = 33;000~- = 2^ ^P- 

111. S. A. E. Determination of Horsepower. — The deter- 
mination of horse power by the S, A. E. (Society of Auto- 
motive Engineers) formula is purely an arbitrary calculation 
and is used only in computing the horsepower of automobile 
motors. The S. A. E. method is derived from the brake 
method and has been useful for comparative purposes such as 
the amount paid for license fee, etc. 



HV' = ~^^~^ ii^ which, 



D = bore of cylinders in inches, 
N = number of cylinders, 
2 . 5 = a constant. 



The above formula assumes a piston speed of 1,000 ft. per 
min., a mean effective cylinder pressure of 90 lb. per sq. in., 
and a mechanical efficiency of 75 per cent. Since a gasoline 
motor delivers more power at a higher rate of speed and since 
the average automobile motor has a piston speed of 1,500 
ft. per min., it is evident that the S. A, E. formula is more or 
less of a makeshift. To-day many motors are tested by the 
brake method before leaving the factory. 

112. Indicated Horsepower (i.hp.). — The indicated horse- 
power method is confined almost exclusively to the steam 
engine, although it may be used for internal combustion 
engines. It determines the power developed in the cylinder 
and not the power transmitted. Indicated horsepower will 
always exceed brake horsepower, as it makes no deduction for 
frictional losses. 

In finding i,hp,, it is necessary to make use of work dia- 
grams as illustrated in Fig. 83. The diagrams are taken by a 

^ The thickness of the belt is added to the diameter. 



96 MECHANICS 

special device called an indicator and from the diagrams the 
mean effective pressure is found for the stroke. For the 
actual method of taking indicator diagrams, the student is 
referred to books dealing with steam and gas engines. 

P = mean effective pressure in lb. 

per sq. in., 
L = length of stroke in ft., 
A = Area of piston head in sq. in. 
A^ = Working strokes per min. 



. , PLAN . , . , 

^'^P' = 33:000 ^^ ^^^^^^ 



In double acting engines the head and crank ends must be 
figured separately and added together for the total horse- 
power. 

113. Mechanical Efficiency. — The mechanical efficiency of 
an engine is the ratio of its brake horsepower and indicated 
horsepower. 

^ _ brake horsepower 
indicated horsepower 

Questions and Problems 

1. Define and give an example of power. 

2. Give the general formula for power. 

3. Define (a) horsepower; (6) watt; (c) kilowatt. 

4. State the mathematical relation between the above units. 
6. State three difi'erent methods of determining horsepower. 

6. Describe in detail two difi'erent ways of determining b.hp. 

7. What is the S. A, E. formula? Is it accurate? Why? 

8. State and explain the formula for i.hp. 

9. Why is b.hp. less than i.hp? 

10. How may the mechanical efficiency of an engine be determined? 

11. A man weighing 175 lb. clim))s a ladder through a vertical distance 
of 30 ft. in 1 min. What hp. did he ex])end? 

12. A Goulds triplex pump lifts 40 lb. of water a vertical distance of 
5 ft. in 1 min. Compute power expended. 

13. A hoisting engine (gears 80 per cent, efficient) is used in lifting 
cement to the top of a building 50 ft. high. If 1,000 tons is to be lifted 
through a working day of 8 hr., what is the average horsepower expended? 

14. The difference in tension between two sides of a belt is 50 li). If 
the bt^t is running at the rate of 2,500 ft. per minute, what horsepower 
is being transmitted? Express the answer also in watts and kilowatts 

15. The Cole ''Aero Eight" develops horsepower in excess of 80. 



POWER 97 

If the cylinders are 33^ in. bore, what is the horsepower according to the 
S. A, E. rating? 

16. An automobile motor has an indicated hp. of 35 and a brake hp. of 
30. What is the mechanical efficiency of the motor? 

17. In computing the power of an electric motor by a prony brake, 
the following data was obtained: net force of brake arm on platform 
balance 11 lb.; length of brake arm 14 in.; r.p.m. 1100. Find horsepower 
and kilowatts transmitted by the motor. 

18. A belt prony brake is used in figuring the hp. of a gasoline motor. 
From the following data compute the power transmitted by the motor: 
difference in tension of balances = 100 lb.; diameter of pulley = 2 ft.; 
r.p.m. = 1,500. 

19. In determining the horsepower of the Corliss steam engine in the 
Dickinson High School power plant, a group of students reported the 
following data: mean effective pressure = 30 lb. per sq. in.; length of 
stroke = 24 in.; diameter of piston = 18 in.; diameter of piston rod = 4 
in.; r.p.m. = 200. Compute the indicated horsepower of the engine. 
(The area of the piston rod must be subtracted from the area of the 
piston on the ''crank end.^') 



CHAPTER XVI 
ENERGY 

114. Energy. — Energy is the ability to do work. A body 
possessing energy may or may not do work. For instance, a 
stick of dynamite possesses a large amount of energy, but the 
energy will remain latent unless the dynamite is exploded. 
A pound of coal contains over 10,000,000 ft.-lb. of energy. No 
work is done, however, unless the coal is burned and the 
heat energy transformed into mechanical energy, as in the 
steam power plant. Energy should never be confused with 
force or power; they are entirely different terms. 

115. Fixed Energy. — Energy stored up in a body is called 

fixed energy. Food contains stored 

^-^^=^ — ^^^ energy from which the body derives 

its energy and ability to do work. 

Gasoline contains stored energy, every 

pound furnishing about 15,000,000 ft.-lb. 

Fixed energy of this kind is determined 

by burning the substance and measuring 

the heat energy given off during the 

/g' \ combustion. 

; ^ Ground Many bodies possess energy on account 

.,,../../.,..,,///.///.//// ^f ^^^:^^ j^^^Hi^^^^ This form of fixed 

Fig. 86. — IF has 1,000 • n i i j- i n^u 

ft.-lb. of potential energy, energy IS called potential energy. The 

pile driver furnishes a good illustration. 
The heavy weight, or driver when released, will fall toward 
the earth and, on striking an object, will accomplish work. 
A body weighing 100 lb. and suspended 10 ft. above the 
earth has 1,000 ft.-lb. of potential energy (see Fig. 86). 
In order to elevate the body it was necessary to do 1,000 
ft.-lb. of work. The potential energy possessed by any 

98 




ENERGY 99 

suspended body is measured by the amount of work necessary 
to raise the body to its position. 

Potential energy = Weight X height = W Y. h 
Potential energy may also be due to elasticity. If a coiled 
spring, when released, is able to exert an average force of 
100 lb. through a distance of 4 in., it is evident that 400 in. -lb. 
of work have been done. It is likewise evident that the spring 
must have possessed 400 in. -lb. of potential energy. The 
valve springs of a gasoline motor furnish a good example of 
potential energy due to elasticity. 

Potential energy = Force X distance = F X d. 

116. Kinetic Energy. — Kinetic energy is the energy of motion. 
A rotating fly wheel or a moving projectile possesses kinetic 
energy. In fact, any moving body possesses kinetic energy, 
because it has the ability to do work. The same amount of 
work will be necessary to bring a body to rest as was neces- 
sary to set the body in motion. If a motor boat has 50,000,- 
000 ft.-lb. of kinetic energy, 50,000,000 ft.-lb. of work will 
be done in bringing the boat to a stop. 

Kinetic energy = ^M TF = weight of the body in lb., 

^9 { V = velocity per sec. m ft., 
ft. lb., in which y g = 32.16, acceleration of gravity. 

Kinetic energy = 1/2 \ M = mass of the body in grams, 
MV^ ergs, in which [ V = velocity per sec. in cm. 

117. Why a Body Possesses Energy. — Every body possessed 
of energy has, at some time, had work done upon it. A com- 
pressed spring has potential energy, due to the fact that work 
was done in compressing the spring. A steam shovel, sus- 
pended above the ground, possesses potential energy, due 
to the fact that a certain amount of work was done upon the 
shovel in lifting it. Similarly, a motor cycle, coasting along 
a level stretch, owes its kinetic energy to the fact that work 
was done in imparting to it the momentum. Work, it will' 
be seen, consists of taking energy from one body and giving it 
to another body. 



100 MECHANICS 

118. Transformation of Energy. — If the weight shown in 
Fig. 86 is released, it will have 1,000 ft. -lb. of kinetic energy 
on striking the ground. During the fall the potential en- 
ergy was being converted into kinetic energy. At any 
time during the downward course of the weight, the weight 
possessed 1,000 ft. -lb. of energy, that is, the sum of the poten- 
tial and kinetic energy was constant. Starting with poten- 
tial energy alone, we see a gradual loss of potential energy 
and a proportionate increase of kinetic energy until, on strik- 
ing the ground, the body has entirely lost its original potential 
energy and possesses a like amount of kinetic energy. 

Other transformations of energy may be noted such as: 
(1) Transformation of heat energy into mechanical energy, as 
in the steam engine; (2) transformation of mechanical energy 
into electrical energy, as in the steam-electric plant; and 
transformation of mechanical energy into heat energy, illus- 
trated in stopping an automobile by means of brakes. Other 
examples might be given, but these are sufficient to show that 
one kind of energy may be replaced by another. 

119. Conservation of Energy. — According to the law of 
conservation of energy, the amount of energy in the universe is 
constant. In other words, energ}^ may neither be created 
nor destroyed. We have seen, however, that energy may be 
transformed from one kind to another. Useful energy, which 
disappears as such, reappears in some other form and is 
called dissipated energy. The electrical energy of an incandes- 
cent light bulb is transformed into light and heat energy. Since 
the heat energy of no use, it is called dissipated energy. Yet 
in no way has the original electrical energy been destroyed. 
Likewise the heat and light onorgy was not created; it resulted 
from the transformation of th(^ (^lectrical energy. 

120. The Mechanical Equivalent of Heat. — In order to 
measure the amount of heat possessed by a body, a unit 
known as the British thermal unit is employed. The British 
thermal unit (B.t.u.) is the amount of heat necessary to raise 
1 lb. of ivater 1 degree Fahrenheit in temperature. If 10 11). of 
water are raised 4° F'ahrenheit in temperature, then 40 B.t.u. 



ENERGY 101 

of heat energy was necessary to effect the rise. It has been 
found by experiment that there is a strict relation between 
work done and heat generated. It has been determined that 
780 ft. -lb. of work always result in the appearance of 1 Bi.u. 
of heat; or, conversely^ that 1 B.t.u, of heat will produce 780 
ft.-lb. of work. This relation was discovered by James Pres- 
cott Joule/ in whose honor the joule was named. 

^ James Prescott Joule (1818-1889). Famous English physicist and 
chemist. Determined the mechanical equivalent of heat and estab- 
lished the law of conservation of energy. Made many valuable experi- 
ments in electricity and magnetism. 

Questions and Problems 

1. Define energy. Explain why a body possesses energy. 

2. What is fixed energy? Give examples. 

3. What is potential energy? Give examples of potential energy 
due to (a) position; (6) elasticity. 

4. How is the energy in a pound of coal measured? 

5. Give the formulas for potential energy. 

6. What is kinetic energy? Give examples. 

7. Give the formulas for kinetic energy. 

8. Give several illustrations of the transformation of energy. 

9. State and discuss the law of conservation of energy. 

10. What is meant by the mechanical equivalent of heat. Discuss 
%lly. 

11. A mine cage weighing 500 Kg. is suspended at the top of a shaft 
100 m. deep. What is its potential energy with respect to the bottom of 
the shaft? 

12. How much kinetic energy will the cage in the previous problem 
have on striking if the cable breaks? 

13. A coiled spring, when released, exerts a pressure of 50 lb. through 
a distance of 3 in. What potential energy did it possess? 

14. A ton pile driver drops 10 ft. What is the average resistance of 
the pile if it is driven down 2 in? 

15. A truck weighing 2 tons is moving at the rate of 20 m.p.h. What 
force must be applied to stop it in 300 ft. if the engine is shut off? 

16. What is the kinetic energy of a tug weighing 100 tons and moving 
at the rate of 10 m.p.h. f How much work must be done to bring the 
tug to a stop if the steam is shut off? 

17. A projectile weighing 500 Kg. has a muzzle velocity of 1,000 m. per 
sec. What is its kinetic energy? 

18. A baseball we-ghing 5 oz. strikes the catcher's mitt with a velocity 
of 40 ft. per sec. What is the kinetic energy on striking? 



102 MECHANICS 

19. An automobile weighing 3,000 lb. is driven at the rate of 50 m.p.h. 
on a level stretch. The power is shut off just as the car strikes a 3 per 
cent, grade. How far up grade will it go before stopping? Assume no 
loss by friction. 

20. A spike is being driven by a sledge. The sledge weighs 40 lb. 
and strikes the spike at the speed of 10 ft. per sec. If the spike is driven 
in 1 in. at each blow, what average resistance does it work against? 




CHAPTER XVII 
FRICTION 

121. Friction. — Friction is the resistance offered to the move- 
ment of one body over another. This resistance is due to the 
fact that no surface is perfectly smooth. The microscope 
reveals to us that every body, no matter how smooth it may 
seem, is covered with countless 
small projections. When bodies 
are in contact, these projections 
interlock, giving rise to friction. 
Suppose we take as an illustra- -,. o^ t^- 

,^ 1 • 1 1 ^^^' ^'' — Diagram (exaggerated) 

tion two surfaces wnicn nave showing cause of friction. 

been planed and polished with 

the utmost care. When examined under the microscope, the 
surfaces are found to be made up of ^^ hills and valleys'' as 
shown (with exaggeration) in Fig. 87. It will be seen that the 
projections of A sink into the depressions of 5. In order to 
move A horizontally along B, it is necessary to lift the upper 
surface of A until the projections cease to interlock. Lifting 
a body involves work. Since work involves /orc6, it is easy to 
see why force is necessary in overcoming friction. 

122. Advantages and Disadvantages of Friction. — It would 
be impossible for us to do without friction. If there was no 
friction, we would be unable to walk or stand; it would be 
almost impossible for us to pick up objects. Locomotives 
would be useless, as there would be no traction between the 
drivers and rails. Belt driven machines and friction clutches 
would not operate. 

On the other hand, friction is a perplexing problem for the 
engineer. He is constantly endeavoring to eliminate it, 
because it impairs the efficiency of machines, resulting in wear 

103 



104 



MECHANICS 



and dissipated power. Power transmitted through spur or 
bevel gears is cut down markedly by friction. A worm-driven 
automobile, noticeable for silent running, is impaired in 
efficiency by the friction of the worm and worm gear. It is 
evident, from the above discussion, that friction is desirable 
in certain ways and undesirable in others. 

123. Coefficient of Friction. — The fraction representing the 
relation between friction and pressure is known as the coeffi- 
cient of friction, 

static friction 



Coefficient of static friction = 



Coefficient of kinetic friction = 



pressure normal to surfaces 

kinetic friction 
pressure normal to surfaces 




^ ^^^^ > 

Fig. 88. — Apparatus for determining coefficient of sliding friction. 



The coefficient of sliding friction only will be considered here 
and is usually spoken of simply as the coefficient of friction. 
It may be determined by means of the apparatus shown in 
Fig. 88. ABC is an adjustable inclined plane to which may 
be attached various substances such as sheet copper, leather, 
brake linings, etc. Suppose we wish to determine the coeffi- 
cient for carbon and copper. The copper sheet is firmly 
attached to the upper surface of the plane and a 4 lb. carbon 
block is placed on the copper. The plane is now adjusted 
until the block jus^ refrains from moving down the plane. The 



FRICTION 105 

block is gently tapped and the angle 6 slowly increased by the 
fine adjustment J until the carbon begins to move slowly 
and uniformly down the copper sheet. The coefficient of 
friction is equal to the height of the plane divided by the base 
of the plane. 

height 



Coefficient of friction = 



base 



The height and length are measured and are found to be 
6 in. and 24 in. respectively. Hence the coefficient for carbon 
and copper will be 6/24 = .25. Since height/base equals the 
tangent of angle d, it is evident that the coefficient is also 
equal to the tangent of angle 6 (in this case 14°). 

Coefficient of friction = tangent of angle d 

By referring to Fig. 88, the above mathematical relations 
are easily explained. W, the weight of the block, may be 
resolved into two components : Fr, or the force parallel to the 
plane just necessary to overcome friction and Pr. the per- 
pendicular pressure between the surfaces. The coefficient 
is evidently Fr./Pr, By similar triangles it may be proven that 
Fr. is proportional to AC or the height of the plane, and that 
Pr. is proportional to BC or the base of the plane. Hence the 
coefficient is equal to the height/base or the tangent of angle 6, 

124. Laws of Friction. — There are so many elements enter- 
ing into a consideration of friction that it almost impossible to 
formulate any exact laws concerning it. Friction depends on 
materials, pressure, smoothness, amount and kind of lubrica- 
tion, temperature, etc. The following laws are accurate 
enough for ordinary purposes. 

For Solids: 

1. Friction varies with the nature of the surfaces in contact. 

2. Friction is proportional to the perpendicular pressure between the 
surfaces in contact. 

3. Friction is not affected by the amount of the area of the surfaces in 
contact. 

4. Friction decreases somewhat as the velocity increases. 



106 



MECHANICS 



For Fluids: 

1. Except at high speeds, friction is little affected by the nature of the 
surfaces in contact. 

2. Friction is independent of the pressure between the surfaces in contact. 

3. Friction depends upon the area of contact and is proportional to it, 

4. Friction increases with velocity and is very great at high speeds. 

123. Anti-friction Devices. — The modern machine has been 
greatly increased in efficiency through the employment of 
anti-friction devices. The simplest method of reducing friction 
is to make the surfaces in contact as smooth as possible by 
mechanical means, such as planing and polishing. The fric- 
tion is further reduced if the surfaces are lubricated. Bear- 
ings are used to cut down the friction of rotating bodies. A 
bearing is a device by which rotating bodies are held in place. 
They are classed as low^ viediuvi or high duty^ according to the 
load they are able to carry. Bearings are either plain, in 
which one surface rubs against another; hall; or roller. If a 
bearing is designed to take care of a load at right angles to 
the rotating body, it is called a radial bearing; if it takes care 
of a load longitudinal to the rotating body, it is called a thrust 
bearing. Many bearings are a combination of the two. 

126. Plain Bearings. — The accompanying diagram (Fig. 
89) shows the connecting rod bearings of a ^^ Universal Una- 




FiG. 89. — Steam engine connecting rod with plain hearings. 



flow'' steam engine. The rod is made from forged steel and 
fitted with a heavy iron box lined with bal)l)itt at the crank 
end and a bronze box at the crosshead end. Wear is taken 
care of by means of the adjustments shown. Babbitt is 
a light, whitish material and is an alloy of tin, copper and 
antimony. It has a low melting point and is very easy to 



FRICTION 



107 



work. Plain bearings are used almost exclusively on the 
steam engine. 

Figure 90 shows the bearings used on the connecting rod of a 
Cole '^Aero-Eighf automobile and is characteristic of auto* 
mobile connecting rods in general. The rod is joined to the 



-/i/N6 CROOveS 




IBLAOE) 



Fig. 90. — Diagram showing plain bearings used on the connecting rod of an 

automobile. 

crankshaft by a babbitt lined bearing. It is secured to the 
piston by a steel wrist pin forced into place and employs a 
bronze bushing as a bearing. Bronze outwears babbitt but 
is not so easily worked. 

127. Ball Bearings. — Ball bearings are used on magnetos, 
electric motors and generators, automobile transmissions, 





Fig. 91. — Cup-and-cone bail 
bearing. 



Fig. 92. — Annular ball bearing. 



airplane propellers, etc. The cup-and-cone bearing (Fig. 91) 
is designed to stand occasional end thrusts and is adjustable. 
The cup and cone, however, must be set at an angle which 



108 



MECHANICS 



lessens the radial load capacity. The efficiency of this type 
of bearing is greatly reduced when wear occurs. 

The annular ball bearing (Fig. 92) has a heavy vertical load 
capacity, but makes no provision for thrust. If thrust loads 
are present to any extent, thrust bearings must be used in 





Fig. 93.— Fafnir radial ball 
bearing assembled. 



Fig. 94. — Fafnir radial ball bearing 
disassembled. 





Fig. 95. — Fafnir thrust ball 
bearing (flat-race type) unas- 
sembled. 



Fig. 96. — Fafnir thrust ball bearing (self- 
aligning type) unassembled. 



addition. The annular type of bearing is not adjustable and 
must be replaced after wear. It is more correct in design 
than the cup-and-cone bearing and is preferred for that 
reason. 

Fafnir ball bearings are representative of the best and are 
here selected for description. The student should examine 
the illustrations above carefully. 



FRICTION 



109 



Fafnir bearings are made of an alloy of high carbon, chrome 
steel, hardened by a special heat treatment and quenched in 
oil to insure proper density and hardness. The raceways are 
very accurately ground and the balls are guaranteed round and 
true to 1/10,000 of an inch. 

128. Roller Bearings. — Roller bearings have proven very 




Inner race 
Fig. 



97.- 



Roller assembly Outer race 

Hyatt roller bearing unassembled. 





Fig. 98. — Timken roller bearing 
partly unassembled. 



Fig. 99. — Front wheel of an 
automobile mounted on Timken 
bearings. 



satisfactory, particularly for medium and heavy loads. The 
Hyatt and Timken bearings are selected for study on account 
of their wide use. 

The Hyatt bearing (Fig. 97) is used for automobiles, mine 




110 MECHANICS 

cars, line shafts, refrigerating machines, printing presses, etc. 
It consists of flexible, hollow rollers of heat-treated alloy steel 
wound helicalty, and an inner and outer race. If shafts are 

sufficiently hard, the inner 
race may be dispensed 
with. The inner race is 
made from seamless tubing 
of alloy steel, carburized 
and heat treated. The 
outer race is of the same 
material and is either split 
or solid. For heavy work, 
the solid type should al- 
ways be used. The rollers 
Fig. 100. — Timken bearings used on a are held in place by heavy 

worm-drive truck. v • x j x xu j 

bars riveted to the end 
rings. The bars and rings constitute the cage, Hyatt bear- 
ings make no provision for thrust. 

The Timken bearing (Fig. 98) has a ribbed cone of carbon- 
ized, electric steel, with outside taper; tapered rollers of electric 
alloy steel, heat-treated; a tapered cage or roller retainer of 
sheet steel; and a cup or outer race of electric steel, with inside 
taper. It will be observed that this bearing is a combination 
thrust and radial bearing; also that its peculiar construction 
enables wear to be taken up easily. 

Timken bearings are used extensively^ for automobiles at 
points of hard service such as rear and front wheels, differential, 
pinion or worm and transmission. Figures 99 and 100 show 
installations of Timken bearings on the automobile. 

129. Lubrication of Bearings. — The best lubricant for 
bearings is a neutral mineral oil or grease, entirely free of any 
acid, alkali or sulphur. Animal and vegetable oils are apt to 
contain acid in sufficient quantities to cause corrosion. A 
good lubricant will not onh^ lessen friction, but will also 
protect the metallic surfaces from the action of the atmosphere 
and aid in the exclusion of dirt and water. A light or medium 
oil is generally used for high speeds, while liquid grease or 
heavy machine oil is used for slow speeds. 



FRICTION 111 

Questions and Problems 

1. Define friction. 

2. Show, using diagram, wliy force is necessary in overcoming friction. 

3. Discuss the advantages and disadvantages of friction. 

4. What is static friction? Kinetic friction? 

5. What is meant by the coefficient of friction? Describe a method 
of determining it. 

6. State four laws for friction between solids; four laws for fluid 
friction. 

7. State the various methods employed in reducing friction. 

8. What is a bearing? Radial bearing? Thrust bearing? 

9. Describe typical plain bearings and give concrete illustrations 
showing where they are used. Discuss the relative advantages of babbitt 
and bronze. 

10. Describe the ball bearing. For what kind of work is it used? 

11. What is an annular bearing? Cup-and-cone bearing? 

12. Describe the Hyatt roller bearing. What is the advantage of the 
spiral rollers? 

13. Describe the Timken roller bearing. What is the distinctive 
feature of the Timken bearing? 

14. A force of 6 lb. is necessary to draw a sled weighing 50 lb. along a 
horizontal surface. Find coefficient of friction and work done if the sled 
moves 50 ft. 

15. The coefficient of friction for oak on leather is .30. What force 
is necessary to pull an oak block weighing 7 lb. slowly and uniformly over 
a level, leather surface? 

16. If it takes a force of 10 lb. to slide a piece of ice across the floor and 
the coefficient of friction is .06, what is the weight of the ice? 

17. An inclined plane has an upper surface of dry agate. The plane is 
adjusted until a steel block weighing 10 lb, slides slowly and uniformly 
down the plane. Find the coefficient of friction for steel on dry agate, 
if the plane is 5 ft. long and 1 ft. high. 

18. The experiment was repeated with oiled agate and the coefficient 
of friction was found to be .107. What was the height of the plane? 

19. An iron block is hauled along a level stone surface by a force of 
200 lb. acting at an angle of 5° to the horizontal. If the coefficient of 
friction is .50, what is the weight of the block? 



CHAPTER XVni 
SIMPLE MACHINES 

130. The Machine. — The machine is a mechanical device 
designed to do work advantageously, A machine does not 
possess energy in itself, but must receive energy from some 
outside source. The original energy, after various trans- 
formations and dissipations, will be delivered in part to some 
special place where the work is to be done. It should be 
borne in mind that no energy is ever destroyed^ but simply 
disappears as useful energy. The great problem in connec- 
tion with machines is to reduce all operating losses as far as 
possible; that is, to work for a higher efficiency by causing the 
output to approach the input more closely. 

131. Simple Machines. — Simple machines are divided into 
six general classes, according to the principle upon which they 
work; namely, the lever, pidley, wheel and axle, inclined plane, 
screw and wedge. As will be seen later, there are in reality 
only two different classes, namely, the lever and inclined plane. 
The pulley and wheel and axle are modified levers, and the 
screw and wedge are modified inclined planes. All machines 
operate on the principle of the lever or inclined plane or a 
coml)ination of the two. It will be assumed, in studj^ing 
simple machines, that no friction is present and that the effi- 
ciency is therefore 100 per cent. 

132. Important Definitions. — In discussing machines, we 
shall have occasion to refer repeatedly to input, output, 
efficiency, velocity ratio and mechanical advantage. A proper 
study of machines depends upon a thorough understanding 
of the above terms. 

Inpid. — Inpid is the energy received by a machine from so7ne 
outside source. 

112 



SIMPLE MACHINES 113 

Output. — Output is the energy which a machine is capable 
of delivering and is always less than the input. 

Efficiency. — Efficiency is the ratio of the output to the input. 
It is customary to reduce the decimal thus obtained to per cent. 

^^ . output 

Efficiency = 7 

input 

Velocity Ratio. — Velocity ratio is the distance through which 
the driving force acts divided by the distance through which the 
resisting force acts in the same time. 

Mechanical Advantage. — Mechanical advantage is the ratio 
of the resisting force or load to the driving force or effort. 

MA— '^^^^^^^^0 force _ R 
driving force E 

133. Law of Frictionless Machines. — Provided there was no 
friction, the following law would apply to all machines, 

The driving force (E) X the distance (di) through which it 
acts = the resisting force (R) X the distance (^2) through which 
it acts in the same time. 

E Xdi = RXd2 

134. General Law for All Machines. — Since friction can 
never be eliminated entirely, for practical purposes the above 
law should be amended to read, 

The driving force (E) X the distance (di) through which it 
acts X the efficiency (Effic.) = the resisting force (R) X the 
distance {d^ through which it acts in the same time. 

E XdiX Effic. = RXd2 

Since input is equal to output plus the energy used in over- 
coming friction, the law may also be stated. 

The driving force (E) X the distance (di) through which it 
acts = the resisting force (R) X the distance {d^) through 
which it acts in the same time + friction (Fr.) X the distance 
{ds) through which it acts also in the same time. 

E Xdi= {RXd2) + {Fr. X d,) 

It is essential that the student should thoroughly under- 



114 MECHANICS 

stand the laws given in this paragraph. These laws apply to 
all machines (frictionless or non-frictionless). Any subse- 
quent ''laws'' relating to machines will be found to be simply 
special applications of the general law. 

General Questions Relating to Machines 

1. What is a machine? What is the source of energy in a machine? 

2. What are the six simple machines? Into what two ultimate classes 
may they be divided? 

3. What is meant by input, output, efficiency, velocity ratio, mechan- 
ical advantage? 

4. What is the law of ''frictionless" machines? Is perpetual motion 
possible? Why? 

5. What is the ''general law" of all machines? State the law in two 
different ways. Show that this law applies to "frictionless" machines. 

6. How may the efficiency of a machine be increased? 

135. The Lever. ^ — The lever is a rigid rod or bar designed to 
rotate about a fixed point called the fulcrum. Unless otherwise 
noted, levers will be considered weightless. Levers are 
divided into three classes, according to the relative positions 
of the effort, resistance and fulcrum. A lever will be in 
equilibrium when, 

The effort (E) X the yerpendicidar distance (di) to the ful- 
cnwi = the resistance (R) X the perpe7idicular distance (^2) to 
the fulcrum. 

E X di = RXd. 

136. The First Class Lever. — The first class lever is illus- 



J. 






^f 



Fig. 101. — First class lever. 

trated by a crow bar. A heavy weight may be raised by the 
application of a relatively small force. An examination of 
Fig. 101 shows that the fulcrum is between the effort and 
1 It is suggested that Chap. X be reviewed at this point. 



SIMPLE MACHINES 



115 



resistance and that speed is sacrificed for force. In case of 
rotation, the output will be equal to the input and the v.r. 
and m.a. will be numerically the same. 

137. The Second Class Lever.^The second class lever is 
illustrated by a wheelbarrow. An examination of Fig. 102 



p 



f 



!<-. 



E--IOO^ 



d,-3' 



df-I.S- 



> 



Fxc^/ -/?x^^ 



R=200^ 
Fig. 102. — Second class lever. 



shows that the resistance is between the fulcrum and effort 
and that speed is sacrificed for force. In case of rotation, the 
output will be equal to the input and the vs. and m,a. will be 
numerically the same. 

138. The Third Class Lever. — The third class lever is 
illustrated by the fishing rod. An examination of Fig. 103 



E-lOO^ 
< 



4=/5'- 



4=3^ 



A 



Exdj =/?/^ 



R-50^ 



Fig. 103. — Third class lever. 



shows that the effort is between the fulcrum and the resistance 
and that force is sacrificed for speed. In case of rotation, the 



116 



MECHANICS 



output will be equal to the input and the vs. and m.a. will be 
numerically the same. 

Questions and Problems Relating to the Lever 

1. What is a lever? Describe the different classes of levers and give 
an example of each. 

2. State the law of ''frictionless," ''weightless" levers. Show that 
this law is an application of the general law of machines. 

3. In case the lever is not considered weightless, how would the above 
law read? 



^-.y..^ 4^- 



450 



# 



T 



vj'- 



Fig. 104. 



^ 




K-5-'- 



a 



-rrrr g^ 

K 21'-- 

FiG. 105. 



20^ 

X 



A 



4. State the class of lever illustrated in the following cases: (a) 
drawing a nail with a claw hammer; (6) rowing a boat; (c) turning a 
grindstone by foot-power; (rf) using a pair of shears; (e) lifting a pail of 
water; (/) moving the spark control of an automobile; {g) cranking an 
automobile. 

6. Determine the velocity ratio and mechanical advantage in Figs. 
101, 102 and 103; also the input and output. Assume a rotation of 360°. 

6. A first class lever is 10 ft. long. Where must the fulcrum be placed 
in order that a weight of 100 lb. suspended at one end may be balanced 
by a weight of 50 lb. at the other end? 

7. A second class lever is 10 ft. long. 
Where must a weight of 100 lb. be placed 
in order to be balanced by a weight of 50 
lb. suspended at the end of the lever? 

8. xV third class lever is 10 ft. long. 
What force ai)plied 6 ft. from the fulcrum 
will l^alance a weight of 100 lb. at the oppo- 
site end? 

9. What force applied at E (Fig. 104) will produce equilibrium? 
10. A pressure of 25 lb. is applied to the emergency brake lever of a 
Studebaker automobile. If the resistance arm is 2^4 in. and the effort 
is applied 1332 i^^- from the fulcrum, what is the tension in the brake rod? 
What is the m,a,f o.rJ 



r^-V" 



/8- 



AJ^ 



1 



t 



>f 



Fici. 100. 



SIMPLE MACHINES 



117 



11. Find the tension in rod B for the treadle as shown in Fig. 105. 
What force acts on the pin at C? What is the m.a.f v.r'f 

12. Figure 106 represents a lever safety valve used on stationary 
steam engines. The valve is 2 in. in diameter and weighs 3 lb. Assum- 
ing the lever to be weightless, what weight must be suspended at W so 
that the valve will ''blow" at a pressure of 175 lb. per sq. in.? 

139. Pulleys. — The pulley is a wheel {generally of wood or 
metal) having a grooved circumference and able to rotate freely 
about a fixed axis at its center. If several pulleys are combined 
along the same axis in the same sheath which is fixed, and if a 



V//////////////A V///////////////A 





Fig. 



107.— Single 
pulley. 



fixed 



/00# 

Fig. 108. — Single mov 
able pulley. 




/00# 

Fig. 109.— Single fixed 
and single movable pul- 
ley. 



hke combination which is movable is joined to it by means of a 
rope or chain, we have what is known as a block and tackle. 
In principle the pulley is a rotating lever. It may bring 
about a gain in force or an advantage of direction. Pulleys 
are usually arranged horizontally in the sheath; for con- 
venience, the vertical arrangement is used here. 

The single fixed pulley (Fig. 107) has an advantage of direc- 
tion only, E and R being equal since the tensions Ti and T^. 
are 100 lb. each. The m,a, is 100/100 or 1. E' and 72 have 
the same speed in case of rotation; hence the v.r, is 1. 

The single movable pulley (Fig. 108) is used when a gain in 
force is desired. The supporting tensions Ti and T^ are 
50 lb. each and the m.a. is 100/50 or 2. Since E has twice the 
velocity of i2, the v.r, will be 2. 



118 



MECHANICS 



A single fixed pulley is generally used with a single movable 
piilley[{F\g. 109). The extra pulley enables the effort to be 
applied more conveniently. The in.a. and v.r. are the same 
as in the previous case. 

Figure 110 shows another arrangement of a single fixed 
and single movable pulley. The supporting tensions Ti and 




v/////y//y///////A v////////////////^ V////////////////A 



loo^ 



Fig. 110.— Single 
fixed and single mov- 
able pulley. 





JOO^ 



Fig. 111. — Two fixed 
and two niov!i])le pul- 
leys. 



100^ 



Fig. 1 12.— Three 
fixed and two movable 
l>ulleys. 



Ti are 50 lb. each and the m.a. is 100/50 or 2. E will have 
twice the speed of R and the v.r, will be 2. 

Figure 111 represents two fixed and hvo movable pidleys. 
The supporting tensions Ti, T2, T3 and Ta arc 25 lb. each 
and the m.a. is 100/25 or 4. The v.r. will also be 4. 

Figure 112 represents three fixed and hvo movable pulleys. 
The supporting tensions Ti, To, T3, T^ and T5 are 20 lb. 
each and the m.a. is 100/20 or 5. The v.r. will also be 5. 

From the above illustrations it is evident that the effort 
required to lift a load by means of frictionless pulleys will be 



SIMPLE MACHINES 



119 



equal to the resisting load divided by the number of supporting 

strands. 

resisting load _ R 

number of supporting strands N 



Effort to lift = 



Questions and Problems Relating to the Pulley 

1. Define a pulley. What is a block and tackle? How are com- 
mercial pulleys arranged in the sheath? Upon what principle does the 
pulley operate? 

2. For what purpose is a single fixed pulley used? Single movable 
pulley? 

3. What is the relation between the effort, resisting load and number of 
supporting strands? Show that this is a special application of the general 
law of machines. 

4. A single fixed pulley lifts a load of 200 lb. What effort will be 
required? What is the m.a? The v.rf 

5. A single movable pulley bears a load of 300 lb. What effort will be 
required? What is the m.a? The v.rf 

6. Design systems of pulleys with mechanical advantages of 1, 2, 4 
and 5. 

7. A 1,000 lb. casting is to be loaded into a truck. Design a system of 
pulleys to do the work, assuming that the rope will break at any tension 
over 150 lb. 

8. An ^^I" beam is being dragged along the ground by means of a 
single fixed and single movable pulley. The fixed pulley is attached to a 
tree and the movable pulley to the beam. If three men are pulling with 
a force of 50 lb. each, what force is exerted (a) on the beam? (6) On the 
tree? What is the tension in each strand? 



140. The Wheel and Axle.— The 
wheel and axle (Fig. 113) works on 
the principle of a first class lever. 
The effort (E)^ applied tangentially to 
the rim of the wheel, will produce a 
rotation of both wheel and axle. E 
will descend and R will rise. Repre- 
senting the radius of the wheel by 
ri and the radius of the axle by r2, 
it is evident that E X 27rri = R X 27rr2, 
R X 27rr2 ^ _ ^2 




Fig. 113.— Wheel andjaxle. 

and that E = 



27rri 



= ^Xr7 



120 



MECHANICS 



In Fig. 113, ri is 9 in. and r2 is 3. in. Hence the effort 

3 
necessary to lift the load of 300 lb. is 300 X g or 100 lb. The 

m.a. is 3 and the v.r. is 3. 

Questions and Problems Relating to the Wheel and Axle 

1. Describe the wheel and axle and state the purpose for which it is 
used. 

2. State the mathematical relation of the effort, resistance, radius of 
the wheel and radius of the axle. 

3. vShow that the above relation is an application of the general law 
of machines. 

4. The diameters of a wheel and axle are 24 in. and 4 in. respectively. 
What load may be lifted by an applied force of 50 lb. ? 

6. Compute the output and input in problem 4, assuming a rotation of 
720 degrees. Also find the m.a. and v.r. 

6. The m.a. of a wheel and axle is 10 and the load to be raised is 200 
lb. If the diameter of the axle is 5 in., what is the diameter of the 
wheel? 




Fig. 114. — Inclined plane. 



141. The Inclined Plane. 

The inclined plane (Fig. 1 14) 
is a block whose upper sur- 
face makes an angle of less 
than 90° with the horizontal. 
It is used to raise heavy 
loads, such as rolling a bar- 
rel of sugar up into a door- 
way by means of a plank. 

The effort required will depend on the slope of the plane. 

According to the general law of machines, the effort (E) X 

the length (/) of the plane = the resistance (7?) X the height 

(/?) of the plane. 

EXl=RXh or E = RX J = R X sin 6. 

In the accompanying diagram, the plane is inclined at an 
angle of 30° to the horizontal and is 10 ft. long and 5 ft. high. 
An effort of 50 lb. will be necessary to move the weight of 
100 lb. up the plane. The m.a, is 2 and the v.r. is 2. 



SIMPLE MACHINES 



121 



Questions and Problems Relating to the Inclined Plane 

1. What is an inclined plane? For what purpose is it used? 

2. What relation exists between E, R, h and U Between E^ R and 
sin e? Show how these relations are derived. 

3. An inclined plane is 20 ft. long and 2 ft. high. How heavy a load 
can be lifted by a force of 100 lb. applied parallel to the plane? How 
much work is done upon the load and how much work is done by the 
force if the load travels the length of the plane? 

4. A plane is inclined at an angle of 45°. What weight may be lifted 
by a force of 50 lb. acting parallel to the plane? 

5. An automobile weighing 3,000 lb. is being towed up a 2 per cent, 
grade. Neglecting friction, what is the tension of the tow rope? 

142. The Screw. — The jack screw (Fig. 115) is a modified 
incHned plane. A piece of paper 
cut into the shape of an inchned 
plane and wound around a pencil 
will represent the threads. The 
jack screw is used for such pur- 
poses as raising buildings and will 
not reverse when the applied 
force is removed. Assuming a 
complete rotation of the hand 
lever, it follows that the effort 
(E) X 27rr = the resistance (R) 
X the lead of the screw. The 
lead is the distance the screw 
advances during one rotation. For single-thread screws it is 
the same as the pitch (distance between two adjacent threads). 

E X 2Trr = R X lead of screw 

In Fig. 115, the radius^ of the hand lever is 2 ft. and the 
lead is ^ in. If the load (including the moving part of the 
jack) is 1,000 lb., the effort to hft is found as follows: 




Fig. 115. — Jack screw. 



E = 



R X lead 1,000 X M 



= 3.31 lb. 



2irr. 2X7rX24 

The m.a. is 302.1 and the v.r. is 302.1. 

' Radius of the hand lever is measured from the center of the jack to 
the point where the effort is applied. 



122 MECHANICS 

Questions and Problems Relating to the Screw 

1. Describe the jack screw. For what purpose is it used? 

2. Why does the jack screw^ not reverse? 

3. What is meant by the lead of a screw? Pitch? 

4. A jack screw has 4 threads to the inch. If the radius of the hand 
lever is 3 ft., what load may be lifted by a force of 75 lb. applied at the 
extremity of the lever? W^hat is the m.al The v.rl 

5. A small building weighing 8,000 lb. is being raised by means of 4 
jacks, one at each corner. If the radius of the hand lever is 24 in. and the 
pitch of the threads is K in., what effort is necessary at the extremity of 
each lever. What is the m.a. and v.r. of each jack? 

143. The Wedge.— The wedge (Fig. 116) is simply two 
inclined planes laid base to base. It is 
used in splitting timbers and rocks, launch- 
ing ships, raising large weights through 
short distances, etc. All cutting tools are 

Fig. IIG.— Wedge. wedgCS. 

In actual practice, friction is such an 
uncertain quantity, that any law in regard to the wedge is 
only a poor approximation. 

Questions and Problems Relating to the We^ge 

1. Describe the wedge. For what purpose is it used? 

2. In splitting a piece of very hard timber would you use a wedge of 
large or small angle? Explain. 

3. In actual practice, the law of the wedge is only approximate. 
Why? 




CHAPTER XIX 

PRACTICAL STUDY OF MACHINES 

144. Efficiency. — It is assumed that the student has 
acquired, by this time, a satisfactory knowledge of the 
so-called simple machines. Hereafter, machines will be dis- 
cussed from a practical point of view. It is impossible, in this 



1 . 




-R 


^#^^'1 




m 




Wi 


^^^^^H^^^^^^^^^^I^^^^^^^Kl 


1 
1 




1 




. M 


nil 

1 !■! 


K-i 




1 1 
t. 1 1 


Pm 




B 






»i1 






i: 



Fig. 117. — Comparison of a Yale differential, screw-geared and spur-geared 

chain block. 

small volume, to examine many of the machines in common 
use. Those selected, however, will be typical and furnish an 
excellent background for further study. 

In Chap. 18, the general law of machines is stated: 

E X diX Effic. = RXd2 
123 



124 MECHANICS 

Since di/dz is equal to the velocity ratio, the law may also 
be stated: 

E X V.R. X Effic. = R XI 

Thus, to find the input of a machine, determine the velocity 
ratio and multiply it by the effort. The output will be equal 
to the resistance times 1 or, simply, the load. The above 
method gives the relative input and output and is independent 
of the actual distance covered by E and R. It is suggested 
that all problems be solved according to this method. 

145. Chain Blocks. — As a hoisting device, the chain block is 
universally used. It is almost indispensable in machine 
shops, factories, warehouses, etc., where heavy loads must 
be lifted frequently. There are three general types in common 
use: the differential block; the screw-geared block; and the 
spur-geared block. 

Figure 117 shows the Yale differential, screw-geared and 
spur-geared chain blocks of one ton capacity. The picture 
shows the maximum load that each man can comfortably 
lift and the distance through which the load can be lifted in 
30 sec. It is assumed that each man pulls 82 lb. The 
student will observe that the spurgeared block has an advan- 
tage both of speed and efficiency over the other two. 

146. Differential Chain Block. — The differential chain 
block employs two sheaves of slightly different diameter in 
the upper block and one sheave in the lower block. The 
sheaves in the upper block are cast together. Over these 
sheaves runs an endless hand chain. Pockets along the rims 
of the upper sheaves prevent slippage. When the effort is 
removed, the friction of the blocks is sufficient to restrain the 
load from running back. The differential is simple, cheap and 
reliable, but has not the durability or efficiency of the screw- 
geared or spur-geared block. 

The velocity ratio of the differential is determined as follows. 
Referring to Fig. 118 and assuming that the upper sheaves 
make one revolution, E then moves a distance of 27r/?. Side 
B of the chain loop is raised 2x72, but side A is lowered 27rr 



PRACTICAL STUDY OF MACHINES 



125 



y//////////////// 



as the chain unwinds. The loop A 5 is therefore shortened 
27rK — 27rr or 27r(i2 — r). The load is raised only half this 
distance or tt (/2 — r). Thus the effort distance divided by the 
load distance equals 2R/R — r. 

Velocity ratio for differential = p ^ • 

Since R and r are proportional respectively to the number 
of chain pockets in the larger and smaller sheaves of the upper, 

block, it is suggested that these values be 
substituted for R and r in the above 
equation. This method eliminates any 
error in measurement. 

In conducting a laboratory study of the 
differential, the velocity ratio is first 
determined. This is done by counting 
the chain pockets in the upper sheaves and 
substituting these values in the formula 
given above. The calculated velocity 
ratio is checked by finding the relative 
distances moved by the effort and load. 

Next it determined, by means of a spring 
balance, the effort necessary to run the 
machine unloaded. A load of 50 lb. is 
now suspended from the load hook and the 
effort to lift recorded. The load is in- 
creased 50 lb. at a time and the corre- 
sponding efforts to lift taken in each case. 
The following figures were submitted by two students. The 
chain block (quarter-ton capacity) was manufactured by the 
Chisholm and Moore Mfg. Co., Cleveland, Ohio. 

VELociTy Ratio 

(a) Chain pockets in larger pulley of upper block = 9. 
(6) Chain pockets in smaller pulley of upper block = 8. 

2X9 

(c) Computed velocity ratio = q _ o = 18- 

(d) Actual velocity ratio (check), 

Distance effort travels 108 in. 




Fig. 



118. — Differential 
chain block. 



Distance load travels 



6 in. 



18. 



126 



MECHANICS 



Efficiency 



Load in 


Effort to 


Velocity 


Input in 


Output in 


Efficiency 


lb. 


lift in lb. 


ratio 


ft. lb. 


ft. lb. 


in per cent. 





2 


18 


36 








50 


11 


18 


198 


50 


25.3 


100 


18 


18 


324 


100 


30.8 


150 


26 


18 


468 


150 


32.0 


200 


34 


18 


612 


200 


32.6 


250 


43 


18 


774 


250 


32.3 


300 


51 


18 


918 


300 


32.6 



1/5 

O 



50 






























y 




























y 


y 












E' 


^ 


-QV^ 


^ 










^ 






40 








^ 






y 


^^ 








30 






y 


^ 










> 


y^ 


I 












/ 










1 ii 


F^^ 


/ 














20 


/ 


/ 






A 


0^'' 


^ 
















/ 








\ 




















10 


/ 




%/> 


y 
























/ 

1/ 


/ 


y 


























^ 


k 






























D 




s'o 




10 







15b 




20 







25to 




300 



40 



32 



^^^ 



IG 



c 
<1> 



Load "in Pounds 
Fig. 119. — Graph showing relation of efficiency and lifting effort to load 
for the differential chain block. 

Figure 119 is a graph showing the relation of the driving 
effort and efficiency to the load. It is customary to locate the 
loads along the ''x'' or horizontal axis and the efforts and 
efficiencies along the ^'y'' or vertical axis. This graph is 
typical for machines and should be thoroughly understood by 
the student. The load-effort curve starts above zero, as some 
effort is required to drive the hoist unloaded. The efficiency 
curve always has its origin at zero, since the efficiency at zero 
load must necessarily be zero. It will be observed that the 



PRACTICAL STUDY OF MACHINES 



127 



effort is directly proportional to the load and that the efficiency 
rises with the load to about 32 per cent, and then. remains prac- 
tically constant, 

145. Screw-geared Chain Block. — The screw-geared block 
is light, compact and especially adapted to portable work. 
It is cheaper than the spur-geared, although it does not possess 
the speed of the latter. It depends upon the worm and worm 
wheel for its mechanical advantage (See Fig. 120). Effort 
is applied by means of a hand chain over a sheave having 
pockets for the chain links. This sheave is a unit with the 




Fig. 120. — Working mechanism of a Yale screw-geared chain block. 



worm shaft which directly operates the worm wheel. The 
worm wheel in turn rotates the load sheaves. The worm 
and worm wheel are of steel and bronze respectively. Parts 
subject to friction run in oil, insuring proper lubrication. The 
block may be reversed but will not ^^run down.'^ 

To compute the velocity ratio, the diameter of the hand 
wheel and the diameter of the load sheaves is first determined. 
The diameter (Di) of the hand wheel is equal to the distance 
between the mid-points of the chain running around it. The 



128 



MECHANICS 



diameter {D2) of the load sheaves is determined in a similar 
manner. Next the number (iV) of teeth in the worm wheel is 
determined by counting the number of revolutions of A 
required to rotate C once. 

Velocity ratio for screw-geared block = y.^ X N, 

In the laboratory, the velocity ratio is found as described 
above and checked b}^ noting how far the hand chain moves 
while the load chain moves 1 ft. 

Next the effort to drive the block unloaded is determined 
by a spring balance. Various loads are now suspended from 
the load hook and the corresponding efforts to lift recorded. 

The following test of a Yale block was made by two students. 

Velocity Ratio 



(a) Diameter of hand wheel (DO = 5.25 in. 

(b) Diameter of load sheaves (D2) = 2.75 in. 

(c) Number of revolutions of A to one of C = 20 

Dl 5 25 

(d) Calculated velocity ratio = ^. X A^ = ^^ 

(e) Distance moved by hand chain = 456 in. 
(/) Distance moved by load hook = 12 in. 

456 
(g) Actual velocity ratio (check) = -r^ = 38. 



.. X 20 
/5 



38.2. 



Efficiency 



Load in 


Effort to 


Velocity 


Input in 


Output in 


Efficiency 


U). 


lift in lb. 


ratio. 


ft.-lb. 


ft.-lb.. 


in per cent. 





1 


38 


38 








55 


4 


38 


152 


55 


36.1 


105 


7 


38 


266 


105 


39.8 


155 


10 


38 


380 


155 


40.7 


205 


13 


38 


494 


205 


41.5 


255 


16 


38 


608 


255 


41.9 



Figure 121 is a graph showing the relation of effort and 
efficiency to the load. An examination of the graph shows that 
the effort is directly proportional to the load and that the effl- 



PRACTICAL STUDY OF MACHINES 



129 



ciency rises with the load, becoming nearly constant around 41 
per cent. Note that the efficiency curve starts at zero and that 
the load-effort curve starts above zero. 



CO 

o 



20 
16 

8 









































1 . 1 
Ffficienc^L/ 










^ 








y 




" 














,^ 


y^ 






/ 


r 














y" 


y 










/ 












'i- _»■ 


/- 


y 










/ 










>■ 


^ 


>" 














/ 
/ 








P 


\i>^ 


















/ 

/ 






X 


y 




















1 


y 


^ 
























y 





























50 



40 60 IZO 160 

Load in Pounds 



eoo 



e40 



40 



30 



20 



10 



IT 

J- 






280 



Fig. 121.- 



Graph showing relation of efficiency and lifting effort to load for 
the screw-geared chain block. 



148. Spur -geared Chain Block. — The Yale spur-geared 
block is very efficient, frictional losses being reduced to a 
minimum. It is made for capacities of from J-^ ton to 40 
tons. One man pulling 82 lb. on the hand chain can lift 
one ton. The lowering of the load is effected by an automatic 
brake mechanism. Referring to Fig. 122, it will be seen that 
the block employs spur gears based on the principle of the 
planetary system. Movement of the hand chain rotates 
pinion A , The pinion transmits its energy to two intermediate 
gears (B and B) diametrically opposed. The intermediate 
gears mesh with the large internal gear D, causing a revolution 
of the pinion cage E to which the load sheave F is attached. 
The design of the block distributes the pressure equally and 
tends to prevent wear in the bearings. 

149. Chain-drive Bicycle.— The bicycle is a speed machine, 
effort being sacrificed in order to obtain a high velocity for the 
rear wheel. It is evident that both the velocity ratio and 



130 



MECHANICS 




Fig. 122. — Working mechanism of a Yale spur-geared chain block. 

mechanical advantage will be less than 1. The bicycle in 
Fig. 123 is intended for practical study. It is firmly mounted 




Fig. 123. — Bicycle mounted for laboratory test. 

and the pedals have been replaced by a wooden pulley equal in 
radius to the length of the pedals, Thus the efforts may be 



PRACTICAL STUDY OF MACHINES 



131 



applied conveniently by means of a stout cord and a scale pan. 
The loads are suspended from the rear wheel in a similar 
manner. 

In a practical test, the velocity ratio is first calculated. The 
circumference of the driving and driven wheel are determined 
and the number of teeth in the front and rear sprockets. 



V.r, for bicycle = 



circumference wooden pulley 
circumference rear wheel 



X 



teeth rear sprocket 
teeth front sprocket 
The velocity ratio is checked by finding the distance the 
effort travels while the load travels 1 ft. 

Next it is determined how much weight must be applied to 
the driving pulley to run the bicycle unloaded. Various 
weights are suspended from the rear wheel and the correspond- 
ing efforts to lift found. 

Velocity Ratio 

(a) Circumference of driving pulley = 47 in. 
(6) Circumference of rear wheel = 77 in. 

(c) Teeth on rear sprocket = 8. 

(d) Teeth on front sprocket = 24. 

(e) Distance effort travels while load travels 36 in. = 7.34 in. 
if) Computed velocity ratio = 47/77 X 8/24 = .204. 

(g) Actual velocity ratio = 7.34/36 = .204. 

Efficiency 



Load in 


Effort to 


Velocity 


Input in 


Output in 


Efficiency 


lb. 


lift in lb. 


ratio. 


ft.-lb. 


ft.-lb. 


in per cent. 





0.13 


0.204 


0.026 








0.25 


1.37 


0.204 


0.279 


0.25 


89.6 


0.50 


2.62 


0.204 


0.538 


0.50 


92.9 


1.00 


5.12 


0.204 


1.044 


1.00 


95.8 


2.00 


10.12 


0.204 


2.064 


2.00 


96.9 


4.00 


20.12 


0.204 


4.104 


4.00 


97.4 


6.00 


30.20 


0.204 


6.160 


6.00 


97.4 



132 



MECHANICS 



Figure 124 is a graph showing the relation of effort and 
efficiency to load for the bicycle. The load-effort line begins 
slightly above zero, indicating that the amount of friction is 
very small as compared with the machines previously studied 
in this chapter. It is evident that the effort is directly pro- 
portional to the load and that the efficiency increases with the 
load to a little over 97 per cent, and then remains constant. 



100 



o 

D- 



itO 












Cf-^ 


.. . 












^ 










tTTii.!encij- 








— 






















32 


f 

1 
























1 

1 






















y 


lA 


1 

I 




















y 


^ 


1 

1 














t 


,y 


y 






\Q) 


1 










\p 


^f^ 












1 










> 












8 








y 


y 


















A 


^ 




















n 


^ 


f^ 



























Z 3 4- 

Load in Pounds 



80 



GO 



4-0 



u 






20 



Fig. 124. — Graph showing relation of efficiency and lifting effort to load for 

the bicycle. 

150. The Jack Screw. — The principle of the jack screw was 
discussed in Chap. 18. In order to approach actual working 
conditions for an efficiency test, the apparatus shown in 
Fig. 125 is recommended. Sufficient loads may thus be 
obtained to make the test very practical. 

An examination of the above diagram shows that the hand 
lever has been replaced by a grooved wheel equal in radius to 
the length of the original torque rod. Various weights are 
suspended from D and the corresponding efforts to rotate the 
wheel are determined by means of a spring balance attached to 
a stout cord running around the wheel. The experimental 
work is conducted as follows. 

First the velocity ratio is determined. This is found by 



PRACTICAL STUDY OF MACHINES 



133 



dividing the circumference of the wheel by the lead of the screw. 

TT 1 '. ,- r ' 1 circumference of drivinq wheel 

Velocity ratio jor jack screw = — -, — -, — ^ ~ • 

lead of screw 

Next the weight and center of gravity of the beam is found ; 
also the weight of the moving part of the jack. It is sug- 
gested that these figures be recorded permanently on the beam 




Fig. 125. — Laboratory apparatus for test of a jack screw. 



A 



r 



-^, 



L 



^. 



>< 



4 



/y 



^, 



Fig. 126. — Force diagram for the jack screw. 

for quick reference and that yard sticks be attached to the 
beam to facilitate the measurement of distances. The beam 
is levelled and 50 lb. is suspended from D, The effort to 
rotate the wheel slowly is found by means of a spring balance. 
The suspended weight is increased 50 lb. at a time and the 
efforts to lift recorded in each case; also the distances di, rfa 
and ds (See Fig. 126). The load may also be varied by moving 
the jack along the beam. This is not so convenient, however. 
To find the total load (L2) raised by the effort (E), the 
following procedure is used. Referring to Figs. 125 and 126, 



134 



MECHANICS 



it will be seen that A in the picture diagram beconies F in the 
force diagram and is considered the fulcrum of the lever. 
The ^^down'' forces are the weight (W) of the lever concen- 
trated at the center of gravity of the beam and the suspended 
weight (TFi)^ The ^^up'' force is the pressure exerted against 
the beam at B and is labelled L on the force diagram. 

Referring to Fig. 126 and applying a suitable law of levers, 
it will be seen that, 

L X di = W X d2 + Wi X {d2 + da) 
L is determined in each case by the above formula. The 

total load (L2) is equal to L plus the weight of the moving 

part of the jack. 

The following experimental work on the jack screw was done 

in the mechanics laboratory of the Dickinson High School. 

Velocity Ratio 

(a) Circumference of hand wheel = 59 in. 

(h) Lead of screw = .50 in. 

(c) Computed velocity ratio 59/. 50 = 118. 

Data 





Weight of 


Weight of 


Suspended 


i Efifort 










beam (PT) in 


jack in lb. 


weight (Wi) 


{E) in 


di in in. 


di in in. 


di in in. 




lb. 




in lb. 


lb. 








1 


23 


36 


50 


18 


12 


44.5 


49 


2 


23 


36 


100 


: 31 


12 


44.5 


49 


3 


23 


36 


150 


! 44 


12 


44.5 


49 




Efficien 


CY 










Total load (L2) 


Input in ft.-lb. 


Output 


: in ft.- 


Efficiency in 






in lb. 




111 




per cent. 



511 

900 

1,290 



2,124 
3,658 
5.192 



511 

900 

1,290 



24.0 
24.6 

24.8 



1 The ''down" force at F is not considered, as its moment is zero. 



PRACTICAL STUDY OF MACHINES 



135 



Figure 127 is a graph constructed from the above figures. 
The low efficiency is accounted for by the excessive friction. 
As in the previous machines, the effort is directly proportional 
to the load. The efficiency remains constant at around 24 per 
cent. The part of the efficiency curve where no figures were 
taken is represented by a broken fine. 



50 










■^. 


.^^ 
















1 


?5 






'$ 


'A 






































> 


X 


1 


c2i 




J 


t 
















y 


z' 






^ 30 
o 




/ 

1 












jjjj 


^i^ 


V 














f 










^ 


y 












"t20 

<4- 


/ 
/ 










> 
















CD 

10 iH 

UJ 


/ 
/ 








y 


<^ 


















10 


/ 




y^ 


y 

y 






















5 


/ 
1 


y 


/ 

























Fig. 127.— 


(--- 




























8 

to load for 



Gra 


ph i 





4C 
ang 



rela 


6C 

i-tioi 
the 


)0 

3cl ir 
1 of 
jacl 


8C 
1 Po 
effic 

: SCI 




jnds 
dene 
•ew. 


10 
y a 


dO 
ndl 


iftin 


)0 
g ef 


14C 
fort 



151. The Automobile Transmission. — Since a gasoKne 
motor car dehvers its maximum power at high speeds, it is 
absolutely essential that some device be provided for varying 
the velocity ratio between the crankshaft and the rear axles. 
This is effected by a special set of spur gears called the trans- 
mission or change gears. 

Figure 128 shows a typical transmission. It consists of a 
main shaft and countershaft, each carrying spur gears and 
mounted on roller bearings. 

The main shaft carries three gears: (1) the main drive gear 
which receives the power from the motor; (2) the high and 
intermediate sliding gear; and (3) the low and reverse sliding 
gear. The sliding gears are arranged to move along their 
shaft by means of a shifter fork, but the shaft and gears must 



136 



MECHAXICS 



rotate together. A shaft along which gears can sHde and 
which rotates as a unit with the gears is often called a spline 
shaft. 

The countershaft carries four gears, each solid with the 
shaft: (1) the countershaft drive gear; (2) the countershaft 
second speed gear; (3) the countershaft first speed gear; and (4) 
the countershaft reverse gear. 

The neutral position of the gears in P^ig. 128 allows the 



^l^h and Inferrnediaie 
Sliding Ge/^r 



Main Dnve Gear 




Counter 
Shafh. 



Counfer... 
Shaft 
Drive 
Gear 



Second 

Speed^.^ 

Gear 



.'L ow and Reverse 
^ Sliding Gear 



(Main Dn ve Shaft 



^ 



To Final 
" Drive 



^ ^^ /:• 'Reverse Idler 



First 
Speed 
Gear 



^ Reverse Gear 



Fig. 128. — Typical automobile transmission. 



motor to run without transmitting any power to the propeller 
shaft. Since there is a ^4)reak'^ between the main drive gear 
and the high and intermediate sliding gear, the countershaft 
merely idles and the part of the main drive shaft connected 
to the final drive and carrying the sliding gears is cut off from 
the source of power. 

First speed or ^^low/^ used in starting and for heavy loads, 
provides the highest velocity ratio of the forward speeds. The 
low and reverse sliding gear is moved to the left by the 
shifter fork, engaging with the first speed gear on the counter- 



PRACTICAL STUDY OF MACHINES 137 

shaft. Thus the propeller shaft receives its motion from the 
motor by way of the countershaft. 

Second speed or ^^ intermediate^^ is obtained by moving the 
high and intermediate sliding gear to the right, so that it will 
engage with the second speed gear on the countershaft. The 
drive is through the countershaft and the velocity ratio is 
less than in the preceding case. 

For third speed or ^'high/^ the high and intermediate sliding 
gear is moved to the left, it internal teeth meshing with the 
external teeth at the right of the main drive gear. Thus, in 
third speed, the upper shaft rotates as a unit, while the counter- 
shaft idles. It will be seen that the motor crankshaft and the 
propeller shaft have the same number of r.p.m. The velocity 
ratio is less than in second speed. 

For reverse speed, the low and reverse sliding gear is moved 
to the right, engaging with the reverse idler. The reverse 
idler is always in mesh with the reverse gear on the counter- 
shaft and runs idle except in reverse speed. The drive is 
through the countershaft. 

It should be noted that the two sliding gears are never both 
in mesh at the same time. The shifter fork always removes 
the gear in mesh before putting the other gear in engagement. 

Questions and Problems 

1. State a general law of machines involving effort, efficiency, load and 
velocity ratio. 

2. Discuss the construction and advantages of the chain blocks 
described in this chapter. 

3. State how you would compute the velocity ratio of the above 
blocks. 

4. Discuss the bicycle as a machine. How is the velocity ratio 
computed? 

6. Describe a practical laboratory test for the jack screw. How is 
the velocity ratio calculated? 

6. What general conclusions did you draw from a study of the graphs 
in this chapter? Why does an efficiency curve always start at the 
intersection of the "x" and ^^y'' axis? Why does the load-effort graph 
always start above the intersection? 

7. What is the purpose of the automobile transmission? Describe 



138 



MECHANICS 



the transmission in detail, diagramming the position of the gears for 
each speed. 

8. A differential type of chain hoist has 8 and 7 chain pockets respec- 
tively in the sheaves of the upper block. What is the velocity ratio? 
If the efficiency is 30 per cent., how much effort is needed to lift a load 
of 100 lb.? 

9. The velocity ratio of a screw-geared chain block is 38. If the 
efficiency is 40 per cent., what load can be raised by an apphed force of 
10 1b? 

10. In a laboratory test of a chain-drive bicycle (see Fig. 123), a force 
of 25 lb. applied to the driving pulley was necessary to overcome 5 lb. 

applied to the rear wheel. What is the 
velocity ratio if the efficiency at this load 
is 97 per cent. ? 

11. The velocity ratio of a jack screw 
arranged as in Fig. 125 is 200. If the 
efficiency is 25 per cent., what effort is 
necessary to raise a total load of 1,5001b.? 





Fig. 129. 



Fig. 130. 



12. A wheel 12 in. in diameter makes 1,000 r.p.m. It is belted to a 
pulley 6 in. in diameter. The pulley is keyed to a second wheel 30 in. in 
diameter. Assuming a belt slippage of 10 per cent., determine the linear 
speed of a point on the circumference of the second wheel. 

13. The screw of the bench vise in Fig. 129 has 9 threads to the inch. 
The effective lever arm of the hand lever is 5 in. Allowing a 65 per cent, 
loss due to friction, what clamping force is exerted by the jaws, if the 
applied effort is 15 lb. ? 

14. The turnbucklc in Fig. 130 has 12 threads to the inch. The 
threaded ends of the rods arc 2 in. apart. How far apart will thc}^ be 
after 10 complete turns? Give two answers. 

16. Devise a system of pulleys whereby one man pulling not over 75 
lb. can lift a weight of 300 lb. Assume an efficiency of 75 per cent. 
Label the diagram carefully and prove its correctness. 

16. The screw of a % in. bolt has 12 threads to the inch. The effective 
leverage of the wrench used in turning on the nut is 8 in. What is the 
clamping force exerted by the nut, if the effort applied to the wrench is 
10 lb.? Assume a frictional loss of 70 per cent. 

17. A machine delivering 8 hp. loses 500 ft.-lb. of work per second 
due to friction. What is the efficiency of the machine? 



PRACTICAL STUDY OF MACHINES 139 

18. A simple hoisting winch consists of a pinion gear turned by a 
crank and engaging a larger gear. The larger gear turns a barrel around 
which the load chain winds. Crank lever =20 in.; teeth in pinion 
= 24; teeth in larger gear = 72; diameter of barrel = 14 in.; load = 
200 lb.; efficiency = 75 per cent. Find (a) the velocity ratio and (b) 
the effort to lift, if the force is applied perpendicular to the lever. 

19. An automobile weighing 3,000 lb. is being towed up a 2 per cent, 
grade (2 ft. rise in a hundred ft.). Assuming friction to be 15 per cent, 
of the weight, what is the tension in the tow rope? 

20. An electric motor, with a driving pulley 6 in. in diameter, makes 
1,500 r.p.m. This pulley is belted to another pulley 8 in. in diameter. 
The latter pulley is keyed to a shaft terminating in bevel pinion with 
15 teeth and engaging a bevel gear with 75 teeth. The bevel gear turns 
a third pulley 10 in. in diameter. How many r.p.m. does the 10 in. 
pulley make? Assume no belt slippage. 




21. The ''whip on whip" pulley arrangement shown in Fig. 131 is 
used for dragging guns out of ditches, etc. B is fastened to a convenient 
tree, let us say; R to the gun mount; and the snub to a stump. The 
effort is applied at F. Assuming an efficiency of 100 per cent, and 
that the ropes and blocks have no weight, find: (a) the relative distances 
travelled by F, A and R; (b) the tension in each strand, if 8 soldiers are 
exerting a force of 500 lb. at F; (c) the reaction at B. 

22. In studying the transmission of a Studebaker six cylinder auto- 
mobile (gears as in Fig. 128), two students submitted the following 
data: teeth on ring gear driving axles = 52; teeth on pinion driving 
ring gear = 14; teeth on main drive gear = 16; teeth on low and reverse 
sliding gear =31; external teeth on high and intermediate sliding gear 

= 24; teeth on countershaft gears from left to right = 32, 24, 17 and 13 
respectively. Compute the velocity ratio between the motor crank- 
shaft and the rear axles for (a) first speed; (6) second speed; (c) third 
speed; (d) reverse speed. 



CHAPTER XX 
MECHANICAL TRANSMISSION OF POWER 

152. Transmission of Power. — Various methods are 
employed in transmitting power from one point to another. 
The kind of transmission used will depend upon the nature of 
the work to be done. Mechanical transmission is effected 
by means of the following devices: 

1. Shafts 

2. Cowplings 

3. Clutches 

4. Cams 

5. Links 

6. Screw Threads 

7. Chains and Sprockets 

(a) Pulleys and Belts 
(6) Pulleys and Ropes 
(c) Friction wheels 

9. Toothed Gears 

153. Shafts. — A shaft is a rotating bar used in transmitting 
power. Shafts are usually of steel. Alloy steel, such as 
nickel or vanadium steel, is used for heavy work and when 
weight is to be kept down. Marine engine and automobile 
shafts are generally alloys. Hollow shafts have found much 
favor, as they are stronger in proportion to their weight than 
solid shafts. 

A line shaft (Fig. 132) is made up of several lengths of 
shafting coupled together and forming a continuous run. It 
may or may not be a main line shaft. 

The 77iain shaft is a line of shafting attached directly to the 
prime mover (motor or engine). The crankshaft of an auto- 
mobile is a part of the main drive shaft (Fig. 133). 

140 



8. Friction Devices 



MECHANICAL TRANSMISSION OF POWER 141 




Fig. 132. — Line shaft used in a beet sugar factory. 



VALVfS 




CAM 
FOLLOWERS 



PLAIN 

SPUR 

TIMING 

GEARS 



CAM 
SHAFT 



CRANK SHAFT 



Fig. 133. — Working parts of a Ford motor. 




Fig. 134.— Countershaft used in shops and factories. 



142 



MECHANICS 



The countershaft is located between the main shaft and the 
driven machine. It is usually a short section and serves 
principally to effect changes of speed and direction. The 
cam shaft of an automobile (Fig. 133) is an example. The 
Dodge countershaft shown in Fig. 134 is widely used in fac- 
tories and shops. Attention is also directed to the trans- 
mission countershaft in Fig. 128, by means of which various 
velocity ratios are made possible between the motor crank- 
shaft and the rear wheels. 

A secondary shaft is here used to include all shafts except 
main and countershafts. The spline shaft in Fig. 128, is a 
secondary shaft except in third speed. In third speed, it is a 
part of the main shaft. 

A flexible shaft is used in transmitting power at a curved 
distance from the source. The dentist^s drill and boring tools 
for mines, etc., are examples. 

154. Couplings. — Line shafts exceeding 25 ft. in length are 
usually made up of separate sections joined together by 





H-Li 


9i 


i 

1 




r^S^^^^^^k ' '''-^K 




^ 


jpf 




Fig. 135. — Flange fixed coupling. 



Fig. 136. — Solid-sleeve fixed 
coupling. 



couplings. Couplings are also used for short sections as in a 
motor-generator set. 

Fixed couplings are used when a shaft is subject to rotation 
only and is free from sudden and dangerous torques. The 
flange type of fixed coupling (Fig. 135) is generally used on 
large shafts. The solid sleeve fixed coupling (Fig. 136) is 



MECHANICAL TRANSMISSION OF POWER 143 

suitable for small shafts. Other types of fixed couplings are 
omitted through lack of space. 

Flexible couplings are used for shafts liable to poor alignment 
or subject to sudden and dangerous torques. The flange 
flexible coupling (Fig. 137) is used for direct connections, such 



ij 


.^^n 



Coil Spring 



y^ommuUtor Cd«er 




Fig. 137. — Flange flexible Fig. 138. — Flexible coupling of an automobile 
coupling. starting motor. 

as gas engines, turbines, motors, generators, etc. The opposite 
parts are held together by flexible pins instead of rigid bolts. 
The pins are made of tempered steel leaves. Figure 138 
represents an Auto-Lite starting motor used on Chevrolet 
motor cars. The driving pinion, engaging with teeth on the 
flywheel, is connected to the armature shaft by means of a 
coiled spring. Thus the spring absorbs much of the sudden 
initial torque. 

155. Universal Couplings. — Universal couplings are used 





Fig. 139. — Single universal 
joint. 



Fig. 140. — Double universal joint. 



for shafts subject to disalignment, as the propellor shafts of 
automobiles, etc. 

The universal couplings shown in Figs. 139 and 140 are 
constructed with steel bolts fitting into bronze bushings. The 



144 



MECHANICS 



universal coupling is low in efficiency, but absolutely neces- 
sary for certain work. 

156. Clutches. — Couplings which admit of ready disengage- 
ment are called clutches. Clutches are usually of the positive 




Fig. 141. — Common types of jaw clutches. 

type or the friction type, although magnetic clutches have 
been used to some extent. 

The jaw clutches shown above are of the positive type. 
It will be noted that the opposite parts interlock, eliminat- 
ing any possibihty of slippage. 

The friction clutch utilizes fric- 
tion to hold the opposite parts 
together. Friction clutches are 
generally of the cone or the disc 
type, the contracting-band and 
expanding-band clutches having 
become nearly obsolete. 

Figure 142 shows a cone clutch 
such as is used on automobiles. 
Friction is caused by the pressure 
of the clutch facing against the 
metallic balance wheel. The pres- 
sure is due to a very strong coil 
spring. The spring is adjustable, so that it may be com- 
pressed in case of sHppage. By pushing in the clutch pedal, 
the transmission is cut off' from the source of power. The 
cone clutch must be kept free from dirt and should be 




Fig. 142. — Cone clutch used on 
an automobile. 



MECHANICAL TRANSMISSION OF POWER 



145 



given an occasional application of Neat's foot oil or castor 
oil, if the facing is leather. Cone clutches must be re-faced 
eventually. 

The cone clutch is rapidly being superseded by the multiple- 
disc clutch for pleasure cars. The multiple-disc clutch used 
on the Dodge car consists of seven discs held in engagement 
by a heavy coil spring. The driving discs (four in number) 
are carried by six studs projecting from the flywheel. The 
driven discs (three in number) are carried by three studs 
riveted to the clutch spider. The driving discs are faced 
on both sides with wire-woven asbestos fabric. The' driven 




Fig. 143. — Discs from a Dodge multiple-disc clutch. 

discs are unfaced steel. If the clutch pedal is pushed in, the 
spring is compressed and the driving discs can rotate independ- 
ently of the driven discs. The source of power is thus cut off 
from the transmission. Modern practice favors the dry 
disc clutch as it gives a high coefficient of friction. Metal 
against metal gives a coefficient of .15 as a rule; this falls to 
about .07 if the surfaces become greasy. Metal against fric- 
tion material will give a coefficient of .27 dry and .10 oiled. 
Thus it will be seen that metal against friction material tends 
to produce the greater friction. The disc clutch offers a large 
frictional area for a comparatively small clutch diameter. 
Furthermore, it does not ^^grab.'^ Since wear is even, there 
is less danger of slippage and less necessity for renewal of 
facings. 

157. Cams. — The cam is an inclined plane in principle. 
The shape of a cam will depend upon its use. It is used to 

10 



146 



MECHANICS 



operate the valves of an internal combustion engine and to 
time the spark; to operate punch presses, shears, etc. Cams 
require good lubrication, as they work with a great deal of 
friction. They are necessary in certain cases, but should be 
avoided whenever possible. The cam mechanism for lifting 
the valves of a Ford motor is shown in Fig. 133. The six- 
sided cam in Fig. 144 is used on a Chalmers six cylinder 
motor for timing the ignition. Other cars use a similar 
arrangement. The point of the cam strikes the contact point 



CONTACT f>OlNT lEvCR 



POINT or CAM 




CONTACT POINT CAP 



Timer cam 



Fig. 144. — Automobile spark timer operated by a cam. 



lever, interrupts the flow of current in the primary circuit by 
separating the contact points, inducing a ''jump'' spark in the 
secondary circuit (and hence in the spark plug) at the proper 
instant. 

158. Links. — Power is often transmitted by means of 
link work, that is, by means of levers, cranks, rods, etc. Links 
may be used for all kinds of loads. The following are common 
examples of link action: the foot-treadle of a lathe; the con- 
necting rod of a steam engine; the valve rod of a Corliss engine, 
etc. Link work should be used wherever possible, since it 
possesses the least friction of any kind of transmission. 

159. Screw Threads. — The projections left in cutting 
helical grooves about a cyHnder are called threads. External 
projections form male threads; internal projections form 




MECHANICAL TRANSMISSION OF POWER 147 

female threads. Thus a bolt is said to have male threads and 
a nut to have female threads. A screw is single^ double or 
triple threaded according to the number of parallel threads 
which it has. The lead of a screw is the linear distance that 
the male or female part advances during one complete turn. 
The lead is often called pitch in connection with single threaded 
screws and is equal to the distance between the centers of two 
adjacent threads. 

Screw threads are used for two purposes : binding or fasten- 
ing of parts and transmission 
of power. Transmission 
screws are generally square v!^^ "m (d) 

thread or acme thread, as Fiq. 145.— Types of screw threads: 
such threads operate with Square thread (a) ; Acme thread (h) ; 
1 c ' .' ,1 .1 1 • 1 American or Sellars thread (c) ; English 

less friction than other kinds, or Whitworth thread (d) . 
The square thread screw is 

used on the jack screw, hand presses, etc. The acme thread 
is similar in construction, but the sides slope slightly. The 
acme screw is stronger and is used, for example, on the lead 
screw of a lathe. It is especially desirable for work in which 
the screw engages with a split nut. The split nut allows the 
connection to be broken at will. Screws of the '^F^' type 
are used for binding, as they operate with a large amount of 
friction. 

160. Chains and Sprockets. — Chains are desirable for 
transmitting power short distances when slippage is to be 
avoided. They are extensively used for bicycles,* automobile 
trucks, electrically driven machine tools, etc. Velocity 
ratios of 1-8 and linear speeds up to 4,000 ft. per minute are 
possible. The sprockets, as a rule, should not have under 16 
teeth and the distance between the sprocket centers should 
not be less than 13-^ times the diameter of the larger sprocket 
or more than 12 ft. As all chains stretch, provision must be 
made for shortening the chain or moving the sprockets farther 
apart. The pressure on the rivets or pins of a chain should 
never exceed 700 lb. per sq. in. Practical tests of a chain 
driven bicycle have shown an efficiency of 98 per cent. Block, 



148 



MECHANICS 



roller and Renold {silent) chains are the ones most commonly 

used. 

The Whitney block chain is shown in Fig. 146 and the 
patent connecting link for the same is shown in Fig. 147. 




Fig. 14G. — Block chain. 





Fig. 147. — Patent connecting link. 



'jr^ m^w^\ dmLm 




Fig. 148. — Roller chain (detachable type) 



m i»^M(iliiflKL t 




Fig. 149.— Holler chain (riveted type). 



Block chains should not exceed 800 ft. per minute in linear 
speed. 

The Whitney roller chain (detachable type) is shown in 
Fig. 148. The riveted type is shown in Fig. 149. It should 
be noted that the detachable type provides for easy removal 
or separation of links. Attention is also directed to the bush- 



MECHANICAL TRANSMISSION OF POWER 



149 



ings surrounding the rivets or pins. This type of a chain will 
operate successfully up to a linear speed of 1,000 ft. per minute. 




o w o^ ^- > 












^ ^^P^' 



Fig. 151. — Link-Belt silent chain 
showing perfect fit to sprocket. 



Fig. 150. — Link-Belt silent 
chain. 



The dleni chain (Fig. 150) 
is manufactured by the Link- 
Belt Co., and is suitable for 
light or heavy duty. It is 
made of the best quality steel 
and consists of a series of 
leaves or plates held together 
by case-hardened bushings and 
case-hardened steel pins. 
Linear speeds of 1,500 ft. per 
minute are easily possible with 
this chain. Figure 151 shows 

how perfectly the chain fits on its sprocket, and Fig. 152 
shows a silent chain in actual operation. 

161. Pulleys and Belts. — BelU are extensively used for 
transmitting power up to 60 ft., beyond which rope transmis- 




3Z5 H.E 



Fig. 152. — Double Link-Belt silent 
chain transmitting 325 hp. 



150 



MECHANICS 





Fig. 153. — Open drive belt (top slack). 




Fig. 154. — Open drive belt (bottom slack) 




Fig. 155. — Open drive belt with fixed idler. 




Fig. 15G. — Open drive l)elt with swing idler. 



.kaA-^Sa***-^ 




Fig. 157. — Open drive belt with adjustable idler. 

I 




Fig. 158. — Crossed drive belt. 



MECHANICAL TRANSMISSION OF POWER 



151 



sion is cheaper. They are suitable for low, medium or heavy 
duty work. Leather belts are best, although cotton, hemp 
or rubber is used in damp places. The strongest and most 
durable belts are made of oak-tanned ox hide. Belt tension 
should not exceed 350 lb. per sq. in. of cross section. Electrical 
transmission has rQplaced the belt in the shop to a certain 
extent; yet the belt will always retain its field of usefulness. 
All belts tend to slip or creep, resulting in wear and loss of 
power. By using top slack (Fig. 153) a greater area of contact 
is effected between pulley and belt and slippage is reduced. 




Fig. 159. — Compound drive belt. 




Fig. 160. — Tandem drive belt. 



Figure 154 illustrates bottom slack. If pulleys are not hned 
up exactly, guides are necessary to keep the belt from coming 
off. 

Short belts generally carry a fixed, swing or adjustable idler 
to minimize creep. It will be seen from Figs. 155, 156 and 
157 that this arrangement provides greater frictional area 
between the pulleys and belt. 

When the driven pulley must rotate opposite in direction to 
the driving pulley, crossed drive (Fig. 158) is used. 

Figures 159 and 160 show the compound and the tandem 
drive belt. For further systems of belting the student is 
referred to a more advanced book. 



152 



MECHAXICS 



Below is a photograph of a 30 in. double Ladew '*Fhnt- 
stone^' belt 176 ft. long. This belt, according to the manu- 
facturers, had just replaced another Ladew belt in continuous 
use for 18 years and crippled through accident, not wear. 




Fig. 161. — Ladew oU-iu. belt (tandem dr;ve; carrying 1,000 hp. 

Referring to Fig. 162, J. is a driving pulley and B a driven 
pulley, direction of rotation being shown by arrows. It is 

evident that the actual driv- 
ing tension is largely To. Ti is 
also in tension and tends to 
retard the clockwise rotation 
of B. Hence the total effect- 
ive driving tension will be To — 
Ti. In case there is no slip- 
page, the horsepower transmitted is as follows: 

_difference in tension in lb, X velocity of belt in ft. permin. 
^^^* " 33^)00 

It is generally assumed that a single belt 1 inch wide and 
running 1,000 ft. per minute will transmit 1 horsepower. The 
rule is not exact, however. The following table, prepared by 
the Ladew Belting Co., will be found conservative. 




Fig. 102. 



MECHANICAL TRANSMISSION OF POWER 



153 



Horsepower Transmitted by Single Belts at Various 

Speeds 



Belt speed ft. per 


750 


1200 


1800 


2400 


3000 


3600 


4200 


4800 


5400 


6000 


mm. 






















Width, inches 








Horsepower 






1 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


2 


2 


4 


6 


7 


9 


11 


13 


15 


16 


18 


3 


3 


5 


8 


11 


14 


16 


19 


22 


25 


27 


4 


4 


7 


11 


15 


18 


22 


25 


29 


33 


36 


5 


5 


9 


14 


18 


23 


27 


32 


36 


41 


45 


6 


6 


11 


16 


22 


27 


33 


38 


44 


49 


55 


8 


7 


15 


22 


29 


36 


44 


51 


58 


65 


73 


9 


8 


16 


25 


33 


41 


49 


57 


65 


74 


82 


10 


9 


18 


27 


36 


45 


55 


64 


73 


82 


91 


12 


11 


22 


33 


44 


55 


65 


76 


87 


98 


109 


14 


13 


25 


38 


51 


64 


76 


89 


102 


115 


127 


16 


15 

t 


29 


44 


58 


73 


87 


102 


116 


131 


145 



162. Pulleys and Ropes. — Rope transmission is used exten- 
sively in medium and heavy duty work, for both main and 
individual drive, when the distance between shafts is rela- 
tively great. It is cheap to install and maintain, noiseless 
and very flexible in application. 

Textile ropes are generally made of cotton, hemp or manila 




Fig. 163. — Manila transmission rope. 

hemp. Cotton is very pliable and is preferred for light work. 
Manila hemp is the strongest and more suitable for heavy 
work. Textile ropes are made of strands (usually four and 
never more than six) wound around a central core. A special 
dressing or lubricant is used to overcome internal friction 
and insure long life. Figure 163 shows a four strand Dodge 



154 MECHANICS 

''Firmus'' manila transmission rope. A four strand manila 
rope 1 inch in diameter will support a tension of nearly 9,000 
lb. If the diameter is doubled, it will support a tension of 
over 32,000 lb. 

Wire ropes are used to some extent, chiefly for hoisting 
purposes. The usual form consists of six strands with a hemp 




Fig. 164. — Wire transmission rope. 




Fig. 165. — Rope transmission used in a hydro-electric plant. 

or wire core. Individual strands arc composed of cast steel 
wires with tensile strengths as high as 250,000 lb. per sq. in. 
There arc generally seven or nineteen wires to a strand. By 
making the wires less in diameter and increasing their number, 
greater pliability is obtained. Thus a ninetccn-wire strand is 
more pliable than a seven-wire strand of the same cross- 
section. The wire rope in Fig. 164 has six strands, each strand 
being made up of nineteen wires. 

Two systems of rope driving are in use: the English or 
multiple system and the American or continuous system. The 



MECHANICAL TRANSMISSION OF POWER 



155 



English system provides numerous separate ropes connecting 
the main drive pulley to the various driven shafts, or separate 
ropes connecting the main drive pulley to the main driven 
pulley, the latter operating the distributing shaft. The 
American system employs a continuous rope connecting the 
main drive pulley to the driven line shaft, to which shaft 
the separate units are attached. In case of a broken rope, 
the EngHsh system will not be out of commission entirely; a 
broken rope in the American system necessitates an immediate 
shut down. Figure 165 represents a rope transmission 
installed in a hydro-electric plant. The total power trans- 
mission is about 500 hp. Attention is called to the grooved 
pulleys. The grooves act as guides and also serve to increase 
the friction of the wheel and rope. 

163. Friction Wheels. — Friction wheels are sometimes used 






Fig. 166. — Friction drive formerly used on a Metz automobile. 

for hght and medium duty, in case various speed changes 
are desired.. There are three types of friction wheels: the 
spur, the bevel and the disc. The friction surfaces may be 
of metal, wood, fiber, leather, etc. At best, friction wheels are 
unsatisfactory and have lost much of their popularity. It is 
nearly impossible to maintain an even pressure and there is 
great stress in the bearings. Figure 166 represents friction 
drive applied to an automobile. A steel plate is connected 
directly to the main drive and engages, at right angles, with 




156 MECHANICS 

a fiber covered wheel connected to the final drive. A gear set 
is thus avoided, changes of speed and direction being obtained 
by sliding the driven wheel along the spline shaft on which it is 
mounted. Friction transmission has attained very little 
prominence in the automobile world. 

164. Toothed Gears. — Toothed gears form a very positive 
method of transmitting power from one shaft to another. 
Teeth from one gear engage similar teeth in another gear, 

slippage being avoided. For low 

speeds and rough work, gears may 

^^ be cast from iron. For high 

^ ^ speeds and where vibration is 

Involute gear teeth. i • ui i r 

undesu'ablc, gears are cut from 
steel. Alloy steel is generally used for the purpose. In order 
to cut down chatter, gears are sometimes made of fiber or 
rawhide. Such gears will not stand as much stress as steel 
gears and are used for light work, such as the timing gears of 
a gasoline motor. Most gears employ the involute tooth (Fig. 
167), other shapes having become nearly obsolete. The 
elements of gear teeth may be either straight lines or helical 
(curved) lines. 

165. Classification of Gears. — Gears are classified with 
reference to the elements of the teeth and the relative position 
of the shafts. Following is a general classification of gears in 
common use: 

1. Spur Gears. — Connect parallel shafts; tooth elements 
straight. 

2. Bevel Gears. — Connect shafts whose axes meet if pro- 
longed; tooth elements straight. 

3. Worm Gears. — Connect shafts that are not parallel and 
that do not meet if prolonged; consist of a wheel with curved 
tooth elements in engagement with a worm having continuous 
screw-like tooth elements. 

4. Helical Gears. — Connect parallel shafts; tooth elements 
helical. 

5. Spiral Gears. — Connect shafts that are not paraUel and 
that do not meet if prolonged; tooth elements hehcal. 



MECHANICAL TRANSMISSION OF POWER 



157 



166. Spur Gears. — Spur gears have teeth whose elements 
are straight Hnes. They are used to connect parallel shafts 
that are close together. Spur gears are extensively used for 
automobile transmission, lathes, pumps, hoisting apparatus, 
etc. 

Figure 168 shows the ordinary spur gear in engagement with 
its pinion (small gear). The annular or internal gear in Fig. 
169 is used in the final drive of certain trucks. Attention is 





Fig. 168. — Spur gear and 
pinion. 



Fig. 169. — Annular or internal gear and 
pinion. 



called to the fact that the tooth form of the internal gear is 
reverse in shape to that of the pinion gear. 

167. Bevel Gears. — Bevel gears have teeth whose elements 
are straight Hnes and connect shafts whose axes meet if 
sufficiently prolonged. The axes are usually at right angles, 
although such a condition is not necessary. A gear bevelled 
at an angle of 45° is called a mitre gear. If the teeth of a 
bevel gear are curved, it is called a curved-tooth or helical 
bevel gear. 

The bevel gear in Fig. 170 has straight teeth. The heHcal 
bevel gear (Fig. 171) is commonly used in the final drive of 
most pleasure cars. The effect of the curved teeth is to 
obtain a smoother and quieter drive and to increase the 
number of teeth in contact at a given instant. 



158 



MECHANICS 



168. Worm Gears. — In worm gears the tooth elements are 
screw Hke; the shafts are not parallel and do not meet if pro- 
longed. Worm gearing affords smooth running and high 




Fig. 170. — Bevel gear and pinion. 



Fig. 171. — Helical bevel gear and ^ 
pinion. 




Fig. 172. — Worm and worm gear. 



speed reductions and has been used with great success on 
motor trucks. The worm and worm gear shown in Fig. 172 
is designed for rear axle drive, but can be used for other pur- 
poses if a speed reduction is desired. The particular ' gear 
shown here is phosphor-bronze and the worm is hardened 



MECHANICAL TRANSMISSION OF POWER 



159 



steel. The object of the wide tooth angle is to decrease 
friction. This makes for increased efficiency of transmission 
and longer life for the engaging parts. 

169. Helical Gears. — The tooth elements of helical gears 





Fig. 173 — Helical gear and pinion. 



Fig. 174. — Double helical 
or herringbone gear. 



are curved or helical. Helical gears connect parallel shafts. 
They are stronger and quieter than spur gears of the same 
width. The timing gears of many automobiles are now of the 
hehcal type. 

Figure 173 shows a pair of 
plain helical gears, similar to 
those used for automobile valve 
timing. The double helical or 
herringbone gear (Fig. 174) is 
used for heavier duty. 

170. Spiral Gears. — Spiral 
gears have curved tooth elements 
and connect shafts that are not 
parallel and that do not meet if 
prolonged. High speed reduc- 
tions are possible and the connection of shafts in different 
planes is also brought about. Some authorities make no 
distinction between helical and spiral gears. 




Fig. 175. — Spiral gears. 



160 MECHANICS 

Questions and Problems 

1. State nine general methods for the mechanical transmission of 
power. 

2. What is the purpose of a shaft? Of what materials are shafts 
made? 

3. What is meant by a line shaft? Main shaft? Countershaft? 
Flexible shaft? Secondary shaft? Give a practical example of each. 

4. What is the purpose of a coupling? State the difference between 
a fixed and flexible coupling. When is a fixed coupling used? A 
flexible coupling? 

5. What is a universal coupling? For what purpose is it used? 

6. What is a clutch? For what kind of work is it used? Describe 
the jaw, cone and disc clutch. What advantage has the disc clutch 
over the cone clutch? 

7. What is a cam? For what kind of work are cams used? Why are 
they avoided whenever possible? 

8. What is a link? Give various examples of power transmitted by 
link work. Why are links used whenever possible? 

9. What is meant by a thread? What is a male thread? Female 
thread? What is the difference between pitch and lead? 

10. For what two general purposes are screw threads used? Describe 
four common types of threads. 

11. For what kind of work is chain transmission desirable? Describe 
the roller, block and silent chain in detail. 

12. When is belt transmission used? Why is ''top slack" better than 
''bottom slack"? Show, using diagrams, various belt drives. Describe 
three kinds of idlers. 

13. Given the difference in tension and the belt speed, how is the hp. 
transmitted by a belt determined? 

14. Why is rope transmission much used? Describe the construction 
of the textile and wire ro})e. What is the difference between the Amer- 
ican and English system of rope transmission? 

15. Are friction wheels satisfactory in power transmission? Explain. 

16. What is a gear? How are gears made? What is meant by the 
tooth element of a gear? Show by diagram an involute gear. 

17. Into what five classes may gears be divided? Define each care- 
fully and point out the kind of work to which it is adapted. 



CHAPTER XXI 

FLUIDS 

SECTION I. INTRODUCTORY 

171. The Three Forms of Matter. — Matter is divided into 
three general classes: solids j liquids and gases. The term 
fluid is used to include both liquids and gases. Many sub- 
stances are capable of existing in any of the three forms. 
Water, for example, may readily be changed to ice or steam. 
Some substances are hard to classify, as they seem to occupy 
intermediate positions. In other words, there is an over- 
lapping which makes the classification only approximate. 
Asphalt is ordinarily a solid, but on a very hot day will tend 
to flow and is then considered a liquid. The exact point 
at which matter changes from a solid to a liquid, liquid to a 
solid, etc., is often difficult to determine. 

Solids have a definite shape and a definite volume and may 
be compressed to a certain extent. In solids the cohesive 
force between the molecules is relatively great, causing the 
molecules to maintain a fixed position with relation to each 
other. The latter statement in no way contradicts the state- 
ment, made earlier in the book, that molecules are always in 
vibration. 

Liquids have a definite volume and an indefinite shape 
and are practically incompressible. A quart of water will 
maintain a constant volume, but will vary its shape to meet the 
shape of the containing vessel. The cohesive force between 
the molecules is insufficient to cause a rigid mass, enabling 
the molecules to move about with more or less freedom. The 
incompressibility of liquids is made use of in transmitting 
pressure through them from one point to another. 

Gases have neither a definite shape nor a definite volume. 
There seems to be no cohesive force between the molecules; in 
11 161 



162 MECHANICS 

fact, they behave as if they are in mutual repulsion. A gas 
like ammonia, if introduced into a room, soon expands and 
penetrates all parts of the room. All gases are readily 
compressible. 

172. Density. — Density is the mass of a unit volume of a 
substance. In the f.p.s. system, density is measured in lb./ 
cu. ft. In the c.g.s. system, density is measured in g./cm.^ 
For example, we speak of the density of water as 62.5 lb./ 
cu. ft. or 1 g./cm.^ Density is determined by dividing the 
weight of a body by its volume. 

^ ., 77iass 

Density = — -, 

votume 

173. Specific Gravity. — Specific gravity is a number express- 
ing how many times a substance is as heavy as an equal 
volume some other substance taken as a standard. For solids 
and liquids, water is the standard; for gases, hydrogen is 
usually the standard, although air and oxygen are sometimes 
used. The specific gravity of sulphuric acid is 1.8. This 
means that a cu. ft. of sulphuric acid weighs 1.8 times as much 
as a cubic foot of water. 

weight of a substance 

weight of an equal volume of some substance 
taken as a standard 

174. Pressure Defined. — The term ^^pressure^^ is used so 
commonly in physics that it is very important for the student 
to have a clear understanding of its meaning. 

Pressure = force acting per unit area. 
Pressure is generally given in lb. pei^ sq. in. or in Kg. per 
sq. cm. For example, the manufacturers of the Goodyear 
33 by 4 in. cord automobile tire recommend that the tire be 
kept inflated to a pressure of 70 lb. per sq. in. or 5 Kg. per sq. 
cm. The same pressure may be expressed more briefly as 70 
lb. or 5 Kg., the corresponding units of area being understood. 

Questions and Problems 

1. Discuss the three forms of matter in detail. 

2. Define density. Give examples. 



Specific gravity 



FLUIDS 163 

3. Define specific gravity. Give examples. What substances are 
used as standards in specific gravity? 

4. Show that density and specific gravity are numerically the same 
in the c g.s. system. 

5. Explain clearly what is meant by pressure. 

6. What is the weight of a piece of cast gold (sp. gr. = 19.3) 10 cm. 
on a side? 

7. Two hundred cm^. of annealed copper weigh 1,778 g. What is 
its density? Specific gravity? 

8. A sheet of silver .001 cm. thick and 1 m. square weighs 105 g. 
What is the density of silver? 

9. A vessel is just filled with mercury (sp, gr. = 13.6). If the 
mercury weighs 20 lb., what is the volume of the vessel in cu. ft.? 

10. A cast iron rod (sp. gr. = 7.1) is 1 in. in diameter and 6 ft. long. 
How much does it weigh? 

11. A cu. ft. of cork weighs 15 lb. What is its specific gravity? 

12. If 5 cu. ft. of sulphuric acid weigh 562.5 lb., what is the specific 
gravity of the acid? 

13. How much does a liter of sulphuric acid weigh? Give answer in 
Kg. 

SECTION II. MECHANICS OF LIQUIDS 

175. Pressure in Liquids at Rest. — It is a matter of common 
experience that a pressure exists in all liquids. A wooden post 
thrust down into a body of water 
is forced back so that a portion of 
the post is out of water. Divers are 
unable to go very far down into the 
water due to the pressure to which 
their ears are subjected. Submarines 
rarely operate more than 200 ft. 
below the surface, on account of 
the tremendous crushing effect on Fig. i76. 

the shell. 

Pressure in hquids is due to the action of gravity on the 
various particles of which the liquid is composed. Assuming 
the liquid in Fig. 176 to be at rest, it is evident that any 
selected particle is in equilibrium; that is, the forces acting 
on the particle are balanced and their resultant is zero. If 
such were not the case, the particle would move and the liquid 




164 



MECHANICS 



5 



\ 



Wafer 



would not be at rest. Hence, in a liquid at rest, pressure at 
any point must be equal in all directions. 

It is also evident that the pressure exerted by the liquid in 
Fig. 176 is normal or perpendicular to the surfaces of the 
container ABCD, Were this not true, a component of the 
pressure at any selected point would cause the hquid to move 
along the surface, contradicting our previous assumption that 

the liquid is at rest. Liquid pressure, 
then, always acts at right angles to the 
retaining surface. 

Liquid pressure on a horizontal surface 
is directly proportiorial to the depth of the 
surface beneath the surface of the liquid. 
Referring to Fig. 177, suppose we have 
a rectangular vessel 1 X 1 X 10 in. half 
filled with water. The pressure on the 
bottom will be equal to the weight of 
the water or 1 X 1 X 5 X .OSG^ = .18 
lb. per sq. in. If the vessel is entirely 
filled with water, the pressure on the 
bottom will be 1 X 1 X 10 X .036 = 
.36 lb. per sq. in. Hence by doubling 
the depth, we have doubled the pressure. 
Suppose, further, that the vessel in 
Fig. 177 is filled with mercury (sp. gr. = 13.6). The pres- 
sure on the bottom will then be 1 X 1 X 10 X .036 X 13.6 = 
4.89 lb. per sq. in., or 13.6 times as much as if the vessel 
were filled with water. The pressure is directly proportional 
to the density of the liquid. 

From the above discussion, we may summarize as follows: 

1. At any selected point, the pressure exerted by a liquid is 
equal in all directions. 

2. Liquid pressure always acts at right angles to the retaimng 
surfaces. 

3. The pressure in each layer of a liquid is proportional to 
the depth. 

^ 1 cu. in. water weighs .03(3 lb. 






Fig. 177. 



FLUIDS 



165 



4. With different liquids and the same depths the pressure is 
directly proportional to the density of the liquid. 

Questions and Problems 

1. Show that pressure exists in liquids. 

2. State the laws of liquid pressure. 

3. A vessel 20 in. high is filled with water. Find (a) the pressure on 
the bottom; (b) the pressure 10 in. down from the surface. 

4. Find the pressure in Problem 3 if (a) the vessel is filled with 
mercury; (6) with sulphuric acid. 

5. If the vessel in Problem 3 is filled partly with mercury and partly 
with sulphuric acid, find the pressure on the bottom. 

6. A submarine lies so that a portion of the shell is 200 ft. below the 
surface of the ocean. If the sp. gr. of sea water is 1.03, what is the 
pressure on the shell? 

7. A depth bomb is adjusted to explode at a pressure of 50 lb. At what 
depth will the explosion take place? Solve for salt water. 

8. A reservoir is 70 ft. deep. Assuming a leakage under the retaining 
wall at this depth, what vertical pressure is exerted on the wall? 

9. If the Titanic sank to an average depth of 3 miles, to what pressure 
is the ship subjected? 

176. Transmission of Pressure by Liquids. — In the begin- 
ning of the chapter it was stated that Hquids are practically 
incompressible. This property is made 

use of in transmitting pressure from one 
point to another through liquids, as in 
the hydraulic press, hydraulic jack, etc., 
in which a tremendous force may be 
exerted with only a small applied force. 

Suppose we have a vessel (Fig. 178) 
containing water, fitted with a plunger 
(P) and having orifices (a, b and c) of 
equal cross-section. If the plunger is pushed down, the pres- 
sure of the plunger will be transmitted to the liquid and will 
cause the water to issue from the orifices in streams of prac- 
tically equal length. This shows that the pressure has been 
transmitted through the liquid equally in all directions. 

177. Pascal's Law. — The following law, of great importance 
in hydraulics, was enunciated by Pascal:^ 

1 Blaise Pascal (1623-1662). French scientist, mathematician, philoso- 
pher and author. 



a 

t 


1 


\ 


I 


b c 


1 — 1 


1 P 


H 


?o 


Fi 


G. 17 


8. 





166 



MECHANICS 



SO* 



w 



206q. In. 
C 



£*■ 

I 



?Sq. In. 



0/1 



Fig 



179. — Apparatus for verify- 
ing Pascal's law. 



Pressure exerted on any part of a confined liquid is trans- 
mitted equally in all directions, acts with equal force on equal 
surfaces and at right angles to the surfaces. 

PascaVs Law Illustrated. — C and c (Fig. 179) are two 
communicating C34inders with cross-sections of 20 and 2 sq. 

in. respectively. P and p are 
tight-fitting pistons (considered 
weightless). The space betw^een 
the pistons is filled with some 
liquid as oil. Suppose a force 
of 5 lb. is acting vertically down 
on p. There is, then, a force of 
2.5 lb. per sq. in. being transmit- 
ted to the liquid by the plunger. 
Since the liquid is subject throughout to a pressure of 2.5 
lb. per sq. in., a pressure of 2.5 lb. per sq. in. is transmitted 
to every sq. in. on the piston P. The total force tending to 
lift P is therefore 2.5 X 20 or 50 lb. In order to prevent 
P from moving upward, a weight of 50 lb. must be placed on 
P. In the above case, an applied force of 5 lb. is sufficient 
to balance a load of 50 lb. The resisting and applied forces 
are directly proportional to their respective cross-sections 
Algebraically : 

W :F ::A:a 

176. The Hydraulic Press. — Figure 180 is a simple diagram 
showing the essential parts of a hydraulic press. The press 
is used to compress cotton into bales and for similar work, 
whenever it is desired to secure a large force by the application 
of a small force. The gain in force is effected by making P 
large in comparison with p and by means of the hand lever 
operating the smaller piston. 

The total crushing effect at R (friction neglected) is found 
as follows: 

area p ab 
Since the machine is considered frictionless, the output 
at R will equal the input at E, Hence: 



FLUIDS 



167 



E X downward distance E moves = R X upward distance P 

moves. 

The student should note that, while a gain in force is 




a 



Wai^r y^y^////y/y^ 




i 



\\ 



^yyyy/y/yy7yyyy////^yyy^^y/yy^ 



w 



Wahr 
Fig. 180. — Hydraulic press. 



secured, there is no gain in energy. This is strictly in accor- 
dance with the law of conservation of energy previously studied. 
177. Communicating Columns. — It is a common saying 
that ^^ water seeks its own level/' This is simply another 







B 


_. — — 


A 


Wcfi-ei 


r 





Fig. 181. — Communicating 
columns. 




Fig. 182. — Heights of liquids 
are inversely proportional to their 
respective densities. 



way of stating that water contained in communicating vessels 
will come to rest at a common height. Referring to Fig. 181, 
it will be seen that column A contains more water than column 
B, Nevertheless, the level in each column is the same. This 
is explained by Pascal's Law. Since pressure is dependent 



168 



MECHANICS 



upon depth, it is clear that, in order for the pressure at the 
base of each cokunn to be the same and keep the liquid in 
equilibrium, the liquid in the separate columns must have 
the same depth. 

If liquids of different densities are used, the heights will be 
inversely proportional to the densities of the liquids. For 
instance, Fig. 182 represents a column of water and a column 
of sulphuric acid separated by mercur}^ to keep them from 
mixing. The mercury is first put in and will come to rest 
with both surfaces at a common level. Since it is necessary 
that the level of the mercury shall not change, the pressure of 
the water against the mercury must be the same as the pres- 
sure of the acid against the mercmy. As sulphuric acid is 
1.8 times as dense as water, this condition can be brought about 
only by having the water column 1.8 times as high as the acid 
column. Stated algebraically: 

// :h ::d :D, 

in which H and h are the heights corresponding respectively to 
the densities D and d. 

178. Difference Between Pressure and Weight. — Pressure 
exerted by a liquid on the base of a vessel depends in no way 

upon the shape of the vessel: 
only upon the depth and den- 
sity of the liquid. Suppose 
(Fig. 183) a, b and c to be three 
vessels each with a base of 1 
sq. in., 10 in. high and filled 
with water. The pressure at 
the base of a will be 1 X 10 X 
.036 or .36 lb. per sq. in. The 
weight of the water contained in 
a will be .36 lb. In this case the pressure exerted by the water 
and the weight of the water are the same. Similarly, we find 
that the pressure in b and c is also .36 lb. per sq. in. Inspection, 
however, shows us that the weight of the water in b is less 
than in a ; also, that the weight of the water in c is greater than 
in a. This proves that pressure and weight are different terms. 



fs 


=-= 


io' 


"i^afer 


Y 






Ca) Cb) (c) 

Fig. IS.S. — Diagrams to show differ- 
ence between pressure and weight. 



FLUIDS 169 

Questions and Problems 

1. Why is a liquid better than a gas in transmitting pressure? 

2. Give an experiment to show that liquids transmit pressure equally 
in all directions. 

3. State and illustrate Pascal's Law. 

4. Describe the principle of the hydraulic press. 

5. Explain why ''water seeks its own level." 

6. State the law of heights for communicating columns of different 
densities. 

7. Give an illustration to show that weight and pressure are not the 
same. 

8. Two communicating vessels {A and B) contain water and are 
fitted with pistons. The piston areas are 10 and 1,000 sq. in. respec- 
tively. If a downward applied force of 50 lb. is used on the smaller 
piston, find (a) the pressure transmitted by each piston; (h) the lifting 
effect of the larger piston. 

9. Repeat problem 8, if the pistons have diameters of 4 and 8 in. 
respectively. 

10. The small piston of a hydraulic press, as shown in Fig. 180, is 
2 in. in diameter and the larger piston is 12 in. in diameter, ac is 36 in. 
and ah is 6 in. Find the crushing effect at 7^ if ^ is 40 lb. 

11. Find R in problem 10, if a mechanical efficiency of 75 per cent, 
is assumed. 

12. Find E in problem 10, if R is 1,000 lb. Assume an efficiency of 
70 per cent. 

13. In Fig. 182, if the water is 5 in. deep and the cylinders are 100 and 
10 sq. in. respectively in cross-section, what weight of sulphuric acid 
must be used to keep the mercury level unchanged ? 

181. Total Pressure on the Various Plane Surfaces of a 
Vessel Containing a Liquid. — Pressure has been defined as the 
force acting per sq. unit of surface. It is frequently necessary 
to compute the total force acting against an area exceeding a 
sq. unit in size. This total force is called total pressure and 
must be distinguished carefully from pressure. Total pressure 
is simply the sum of the separate pressures. 

(1) Horizontal Surfaces 

We have seen that the liquid pressure on a horizontal sur- 
face is equal to the unit area X the height of the liquid X 
the density of the liquid. It follows, therefore, that: 



170 



MECHANICS 



The total liquid pressure acting upon a horizontal surface 
equals the area of the surface X the depth of the liquid X the 
density of the liquid, 

(2) Vertical Surfaces 

In dealing with vertical surfaces, we must take into account 
that the pressure against the surface is not constant, but 
increases with the depth. The total pressure will then be the 
area of the surface X the average pressure of the liquid. We 
may state the rule for total pressure as follows: 

The total liquid pressure acting upon a vertical surface equals 
the area in the liquid X the vertical distance from the center of 
gravity of the surface to the top of the liquid X the density of the 
liquid. 

(3) Surfaces Neither Horizontal nor Vertical 

Since liquid pressure always acts at right angles to any 
surface, the rule stated in (2) is applicable to surfaces which 
are not vertical or horizontal. 

The student should note that the law in (2) holds in (1) 
as well. Hence the rule given in (2) for vertical surfaces 
may be used for any plane surface in the solution of problems. 

182. Center of Pressure. — Since pressure against a sub- 
merged surface always acts at right angles to the surface, 




(a) (b) 

Fig. 184. — Center of pressure for a horizontal surface. 

we may conceive of the total pressure as being made up of 
countless parallel forces. The total pressure is the resultant of 
the separate forces. 



FLUIDS 171 

Figure 184(a) represents a horizontal surface acted upon 
by parallel forces. Since the forces are equal, we may assign 
a value of ^'/'' to each. As there are 25 forces, the resultant 
is evidently 25/. If we replace the separate forces by a single 
force of 25/ applied at the center of area -of the surface as 
shown in Fig. 184(6), the effect will|be the same as in Fig. 
184(a). The point at which the separate parallel liquid forces 
acting against a submerged surface may he replaced by a single, 
resultant force, is called the center of pressure for the surface. 
For horizontal surfaces, the center of pressure is always at 
the center of gravity of the surface. 

In case the submerged surface is vertical and rectangular, 
the center of pressure will be lo- 
cated below the center of gravity. 
This is due to the fact that the ' ; 
pressure increases with the depth /^ 
as shown in Fig. 185. Referring : 
to Fig. 185, we may represent the 
separate forces by/, 2/, 3/ and 4/. 
The resultant force is 10 /. It will 

1 1 . , • xi, X xu Fig. 185. — Center of pressure for 

be seen, by mspection, that the ^ rectangular vertical surface. 

resultant force (10/) must be 

applied below the center of gravity to produce the same effect 
as the separate forces. In this case, the center of pressure is ^^ 
of the vertical distance from the top of the surface to the base. 
For surfaces not horizontal, and submerged so that the top 
is in line with the level of the liquid, the center of pressure is 
located as follows: 

1. Rectangular Surfaces. — Center of pressure is % the vertical distance 
from the top to the base. 

2. Triangular Surfaces with Base Horizontal and Vertex at Top. — Center 
of pressure is ^i the distance from the vertex to the mid-point of the 
base. 

3. Triangular Surfaces with Base at Top. — Center of pressure is 3^ the 
distance from the mid-point of the base to the vertex. 

183. Dams and Retaining Walls. — Great care must be 
exercised in the construction of dams and retaining walls. 




172 



MECHAXICS 



The most important feature of a dam or retaining wall is 
the foundation. Poorly constructed foundations allow water 
to leak in, impairing the base and at the same time exerting 
an upward pressure. This may result in a failure of the 
structure. 

Assuming a proper foundation, a dam or retaining wall 
is liable to failure in three waj^s: (1) sliding horizoiitally; 
(2) crushing of materials; (3) overturning. 

1. Stability against sliding is obtained by making the 




/Foc/r 



Fig. 186.— Dam A-B-C-D will 
not overturn around A if moment 
W X di exceeds moment P X ch.. 



187. — Transverse outline of the 
Ashokan dam. 



structure sufficiently heavy and allowing plenty of friction 
at the base. 

2. Stability against crushing is obtained by keeping the 
compression, to which the material is subjected, less than its 
crushing strength. 

3. Stability against rotation is obtained by making the 
structure heavy and thick at the base, causing the stabilizing 
moment to exceed the turning moment. For example, in 
Fig. 186, the dam A BCD if overturned would swing about A 
as a fulcrum. The moment tending to overturn the dam is 



FLUIDS 173 

P X d2. P is the resultant liquid pressure applied at the 
center of pressure and 62 is the moment arm or perpendicular 
distance to the fulcrum A. The stabilizing moment is W 
X di. W is the weight of the dam concentrated at the center 
of gravity and di is the corresponding moment arm. If W 
X di exceeds P X ^2, there will be no rotation about A, 

Figure 187 shows the outline of a transverse section of the 
Ashokan dam, a part of the water system for Greater New 
York. The masonry part is built of irregular rocks cemented 
with poured concrete and faced with concrete blocks. It is 
1,000 ft. long and 220 ft. high. The width of base is 190.2 
ft. and the width at the top is 26.3 ft. 

Questions and Problems 

1. Distinguish between pressure and total pressure. 

2. State the rule for finding the total liquid pressure acting against a 
plane surface. 

3. What is meant by center of pressure? Where is the center of 
pressure located in a submerged horizontal surface? Submerged vertical 
surface? 

4. Why must the base of a dam or retaining wall be constructed 
carefully? 

5. Assuming a proper foundation, mention three causes why a dam 
or retaining wall is liable to fail. 

6. Explain the precautions taken to guard against failure as indicated 
by the answer to Problem 5. 

7. Give a brief description of the Ashokan Dam. 

8. A swimming tank is 50 ft. long, 35 ft. wide and 8 ft. deep. If the 
tank is filled with water, find (a) the weight of the water; (b) the total 
pressure on the bottom; (c) the total pressure on each side; (d) the total 
pressure on each end. 

9. A dam is 100 ft. long and holds back a body of water 2 miles in 
length. If the water is 30 ft. deep, find the total pressure that the dam 
must withstand. 

10. A gate in a dam is 1 m. wide and 2 m. high. The top of the gate is 
4 m. below the water level. Find the total pressure against the gate in 
Kg. 

11. A swimming tank 40 ft. long and 15 ft. wide is entirely filled with 
water. The water is 4 ft. deep at one end and slopes evenly to a depth of 
6 ft. at the other end. Find (a) the total pressure on the bottom; (b) 
the total pressure on each end. 



174 



MECHANICS 



12. A submarine lies so that a certain horizontal section 2 m. square is 
50 m. below the level of the water. Find the total pressure against the 
section in Kg. Density of sea water is 1.03. 

13. A pressure gauge applied to a city main registers 75 lb. How 
high above the gauge is the water in the distributing reservoir? 

14. A covered rectangular tank is 6 ft. long, 4 ft. wide and 4 ft. deep. 
A tube 6 X 4 in. and 24 in. high extends up from the cover. If the tank 
and tube are filled with water, find (a) the total weight of the water; 
(b) the total pressure on the bottom of the tank; (c) the total pressure on 
each end of the tank; (d) the total pressure on each side of the tank; 
(e) the total pressure on each surface of the tube. 

184. Waterwheels. — Waterpower has been used for cen- 
turies. If available, it is both convenient and economical. 





Fig. 188. — Overshot water 
wheel. 



Fig. 189. — Undershot water wheel. 



Many cities owe their existence and growth to natural 
waterfall. Water furnishes the motive power for mills, fac- 
tories, electric plants, etc. Various types of waterwheels are 
in use. Modern practice favors the Pelton and the turbine, 
although the old-fashioned overshot and undershot wheels are 
still used to some extent. 

The overshot wheel (Fig. 188) depends chiefly on the potential 
energy of the water flowing into the buckets; but some energy 
is derived from the impact of the water against the buckets. 
It has been used mostly in mountainous places where the 
fall is good, but the actual volume of water small. The effi- 
ciency of the overshot wheel is often as high as 90 per cent. 
The loss is due to the friction of the bearings, etc., as well as to 



FLUIDS 



175 



the water which misses the buckets or which is spilled from 
them. 

The undershot wheel (Fig. 189) is used in flat regions where 
there is a large volume of water with poor fall. It depends 
entirely upon the kinetic energy of the 
water. Not over 30 per cent, efficiency 
may be expected from a wheel of this 
type. 

The Pelton wheel (Fig. 190) has met 
with much success. It is ordinarily 
used in water motors for household 
purposes or for other purposes where 

a small horsepower is desired. Pelton wheels have been 
built, however, capable of delivering several thousand hp. 
Water at high pressure is directed against the cup-shaped 
buckets and a large part of its kinetic energy is transferred 

to [the wheel. The efficiency of 




Fig. 190. — Pelton water 
wheel. 



rOenemfor 



mfer 






I 



I 

-I 

J 



the Pelton will often run over 80 
per cent. 

The prevailing type of water- 
wheel today is the turbine. The 
turbines at Niagara develop about 




Turbine 
Xase 



iiii!mNi7i^!n!n :z2 



Tail Wa-her 
Fig. 191. — Water turbine. 




^- Wafer 



Fig. 192. — Diagram show- 
ing fixed and movable blades 
of a water turbine. 



5,000 hp. each. Figure 191 shows a typical installation. Water 
enters the penstock and flows down into the turbine case. The 
turbine wheel rotates horizontally in an inner case. The fixed 



176 MECHANICS 

guides (Fig. 192) direct the water against the movable blades 
at the proper angle, rotating the wheel which is connected to 
the electric generator by a vertical shaft. After imparting a 
large part of its energy to the turbine, the water drops down 
into the tail race. The amount of water flowing through the 
case is controlled by means of a gate valve. The maximum 
efficiency will be about 90 per cent. The energy expended 
upon a turbine per second is equal to the product of the 
weight of the water flowing through it per second and the 
height of the water above the bottom of the pit. 

Questions and Problems 

1. Discuss waterpower as a source of energy. 

2. Give the construction of an overshot waterwheel. Explain how it 
gets its energy. What is the maximum efficiency possible with a wheel 
of this type? 

3. Repeat as above for an undershot waterwheel. 

4. Describe the construction and operation of a Pelton waterwheel. 
For what kind of work is it generally used? What is the maximum 
efficiency to be expected? 

5. Describe the construction and operation of a modern turbine power 
plant. Why is the turbine so efficient? How is the energy per second 
expended upon a turbine determined? 

6. The hp. of a Niagara turbine is 5,000. The pit is 136 ft. deep. 
How much water passes through the turbine per minute, if the efficiency 
is 85 per cent. ? 

7. Repeat Problem 0, using efficiencies of (a) 80 per cent.; (b) 90 per 
cent. 

185. Buoyant Force of Liquids. — It is common knowledge 
that bodies weigh less in water than in air. 
A heavy stone may be moved much more 
easily in water than in air. Floating bodies 
lose their entire weight when placed in a 
liquid. The following paragraph will 
explain the buoyant or lifting effect of 

Fig. 193.— Diagram li(iuids. 

showing cause of Referring to Fig. 193(a), we have a rect- 

buovancy. 

angular solid A BCD (side view shown) 
immersed in some liquid as water. The total pressure push- 
ing the body downward is ecfual to the weight of a cokunn 




FLUIDS 177 

of the liquid the size of the volume FADE. The total upward 
pressure at the base of the body is equal to the weight of 
a column of water the size of the volume FBCE, The buoy- 
ant force of the liquid is equal to the difference between 
the total upward and downward pressures or the weight of 
a column of liquid the size of ABCD. Since FBCE = FADE 
+ ABCD J algebraically: 

FADE + ABCD - FADE = ABCD 

From the above, we see that the loss of weight of the body 
is equal to the weight of the liquid displaced. By similar 
reasoning, since liquids are practically the same density 
throughout, it may be proven in Fig. 193(6) that the buoyant 
force will remain the same if the body is farther down in the 
liquid. Hence we may consider buoyancy independent of 
depth. 

The relation between the loss of weight of a body and the 
liquid displaced was discovered by Archimedes^ and is called 
Archimedes' Principle. 

Archimedes^ Principle. — The loss of weight of a body immersed 
in a liquid is equal to the weight of the liquid displaced. 

Since a floating body loses all of its weight when placed in a 
liquid, the weight of the liquid displaced must be equal to the 
original weight of the body. 

Archimedes' principle furnishes a very convenient method of 
determining the specific gravity of solids. For bodies heavier 
than water, the procedure is very simple. The body is 
weighed in air and then in water. The specific gravity is equal 
to the weight of the body in air divided by the loss of weight in 
water. In case of floating bodies, it is necessary to attach 
a sinker to the body of sufficient size to submerge it. The 
following problems will illustrate each method. 

^Archimedes (287-212 B.C.). Greatest mathematician of antiquity. 
Born in Syracuse, Sicily. First determined the value of ir and the area 
of a circle. Discovered the laws of flotation and of the lever. Killed 
by a Roman soldier during the sack of Syracuse. 

12 



178 



MECHANICS 



1000 



1. A piece of wrought iron weighs 10 lb. in air and 8.73 lb. in water. 
What is the specific gravit}^ of the iron? It is evident that the loss of 
weight of the iron in water is equal to the weight of the water displaced. 
Therefore : 

Sp. gr. wrought iron = 10/10 - 8.73 = 7.87 

2. A block of paraffine weighs 2 lb. in air. It is attached to a sinker 
of sufficient size to submerge it. First the block and sinker are weighed 
with the sinker submerged and the paraffine in the air. This weight is 
5 lb. Next the weight is taken with both submerged. This is found 
to be 2.76 lb. Hence the buoyant force of the water on the paraffine or 
the weight of a volume of water the size of the paraffine is 5 — 2 . 76 or 
2.24 1b. Therefore: 

Sp. gr. paraffine = 2/5 - 2.76 = .89 

186. The Hydrometer. — A floating body displaces its 
own weight of a liquid. If the floating body 
is placed in liquids of different densities, the 
volumes displaced in the several cases will be 
inversely proportional to the respective den- 
sities. The hydrometer, an instrument for 
measuring the specific gravity of liquids, is 
constructed to work on this principle. The 
hydrometer shown in Fig. 194 consists of a 
glass tube with two bulbs. The stem and 
upper bulb contain air; the lower bulb is 
loaded with shot so that the apparatus will 
float upright. The instrument shown here 
is for liquids heavier than water. When placed 
in water, the graduation mark 1,000 is even 
with the water level. Since the specific grav- 
ity of water is 1, it is clear that we must 
read the 1,000 graduation mark as 1. When 
placed in sulphuric acid, the graduation mark 
corresponding to the acid level will be 1,800, 
showing that the specific gravity of sulphuric 
acid is 1.8. The lower graduation of the 
h3^drometer here described is 2,000, giving a 
range of from 1 to 2. The hydrometer used for liquids 
lighter than water is usually graduated from 1,000 at the 





i 




\ 

1 


/i? 
) 



2000 



Fig. 194.— Con- 
stant-weight hy- 
drometer. 



FLUIDS 179 

bottom to 600 at the top. When working exclusively with 
certain liquids, special hydrometers are provided which are 
more sensitive. The hydrometer used to test milk is called 
a lactometer; for alcohol, an alcoholmeter, etc. 

Questions and Problems 

1. What is meant by the ''buoyant force of liquids?'^ 

2. State Archimedes' principle. Does it apply to floating bodies? 

3. An aluminum block is 4 ft. long, 4 ft. wide and 3 ft. high. If the 
block is submerged in water, prove that the total upward pressure acting 
on the block minus the total downward pressure on the block is equal to 
the weight of the water displaced. 

4. A piece of wood 4 in. square floats with % of its volume submerged. 
Find (a) the weight of the wood; (b) the specific gravity of the wood. 

5. A loaded auto truck is driven aboard a small ferry. The ferry is 
30 ft. long and 15 ft. wide. If the ferry sinks 1^ ^-y what is the weight 
of the truck? 

6. A body weighs 20 lb. in air and 10 lb. in water. Find its specific 
gravity. 

7. A hollow ball of cast iron weighs 500 g. What must be its volume 
in cm^. in order for it to remain suspended if submerged in water? 

8. What must be the volume of the ball in problem 7, if kerosene 
(sp, gr. = .79) is used instead of water? 

9. A sinker weighs 2,000 g. in air, 1,600 g. in water and 1,588 g. in 
salt water. Find (a) the specific gravity of the sinker; (6) the specific 
gravity of the salt water. 

10. A piece of paraffine (sp. gr. = .89) is 1 ft. square. How far will 
it sink in a salt solution whose specific gravity is 1.03? 

11. A wooden hydrometer 1 X 1 X 30 cm. sinks 27 cm. in water and 
25 cm. in a solution of copper sulphate. Find the specific gravity of the 
solution. 

12. A body weighs 2 Kg, in air. It is attached to a sinker which alone 
weighs 3,000 g. in water. The sinker is just sufficient to submerge body. 
Find the specific gravity of the body. 

13. A piece of wood weighs 2 Kg. in air. It is fastened to a sinker and 
the sinker and wood when suspended together under water weigh 3.5 
kg. The sinker alone in water weighs 6 Kg. Find the specific gravity 
of the wood. 

14. A uniform glass tube 5 ft. long has an outside cross-section of 
4 sq. in. It is loaded with mercury at the closed end. The tube and 
mercury together weigh 15 lb. How many cu. in. of cork (sp. gr. = 
. 25) must be fastened to the tube so that it will float upright with the 
top 1 ft. out of water? The cork must be entirely submerged. 



180 MECHANICS 

SECTION III. MECHANICS OF GASES 

187. Characteristics of Gases. — We have seen in Sect. I 
that gases have neither a definite shape nor a definite volume; 
that they readily admit of compression; and that they tend to 
expand indefinitely if unconfined. We may now state, in 

r addition, that gases follow the laws of Pascal and Archimedes 
and that the pressure exerted by a gas at any point is equal 
in all directions. Gases expand when heated and contract 
when cooled. If a confined gas is heated, it will undergo an 
increase in pressure. If external pressure is applied to a 
confined gas {e.g., the mixture of gasoline vapor and air in a 
gasoline motor cylinder), the volume of the gas will become 
less; if the pressure is removed, the gas will expand to its 
former volume. The above changes will be studied in detail 
later in the chapter. 

188. The Atmosphere. — The earth is surrounded by a vast 
sea of atmosphere. Like liquids, the atmosphere has weight 
and exerts pressure. At sea level the air exerts a pressure of 
14.7 Ib./sq. in. or 1033.6 g./cm^. For approximate work 
we may take the pressure as 15 Ib./sq. in. or 1 Kg./cm^. 
It is supposed, since the pressure decreases with altitude, that 
the extent of the atmosphere is limited. It is believed that 
no atmosphere exists at a distance of about 500 miles above 
the earth. Air is a mixture, containing about 20 per cent, 
by volume of oxygen and 75 per cent, by volume of nitrogen, 
as well as small quantities of carbon dioxide, water vapor, 
etc. At average sea level at a temperature of 0°C., the densit}'- 
of air is .0807 Ib./cu. ft. or .001293 g./cm^. Thus 12 cu. ft. 
of air weigh approximately a pound. A room may easily 
contain a ton of air. If the atmosphere had the same density 
at all altitudes as at sea level, it would extend upward about 
5 miles. This height is called the height of the homogeneous 
atwosphere. 

189. Proof that Air has Weight. — We may prove that air 
has weight by the following simple experiment. Figure 195 
represents a two liter flask attached to the hell glass of an air 



FLUIDS 



181 



pump as shown. First the flask and the air therein contained 
are weighed. The air is then exhausted and the container 
minus the air is weighed. The difference in the weights 
recorded is the weight of the two liters of air formerly con- 
tained in the flask. If the air pump is efficient, the true 
weight of the air will be found very closely and will be approxi- 
mately 2.6 g./liter. 

190. Proof that Air Exerts Pressure. 
It is a matter of every-day knowledge 
that the air exerts pressure. Auto- 
mobile tires, foot balls, etc., depend 
upon air pressure. In constructing the 
1^ Hudson Tubes under the North River, 




^M: 





Fig. 195. — Appa- 
ratus to show that 
air has weight. 



Fig. 196.— Appa- 
ratus to show that 
air exerts pressure. 



compressed air was used to prevent the water from entering the 
tubes while work was under way. The following demonstra- 
tion proves most strikingly that air exerts pressure. A wood 
funnel is fitted tightly into a bladder glass (Fig. 196) and the 
bladder glasg is attached in turn to the bell glass of an air 
pump. Mercury is poured into the funnel and the air is 
withdrawn from beneath it. The air pressure above forces 
the mercury through the minute pores of the wood into the 
catch basin below. This demonstration also proves that the 
wood is porous. 

191. Torricelli's Experiment. — The ancients were well 
aware of the fact that water would rise against gravity in 



182 



MECHANICS 



exhausted tubes. This knowledge was utiUzed in Hfting 
water from wells. The accepted explanation was that 
''nature abhors a vacuum.'' It was discovered early in the 

seventeenth century that water 
would not rise much over 32 ft. 
in an exhausted tube. Galileo's^ 
explanation of this phenomenon was 
that ''nature's horror of a vacuum 
ceased at 32 ft." It is probable, 
however, that Galileo suspected the 
true reason just before he died. 
After Galileo's death, Torricelli,^ his 
pupil-, reasoned since mercury is 
about 13 times as heavy as water, 
that mercury would rise only about 
i^fs as high as water in an exhausted 
tube. In 1643 Torricelli performed 
the following experiment. He pro- 
cured a glass tube about 4 ft. long 
and closed at one end. He then 
filled the tube with mercury, thereby expelling the air (Fig. 
197). Placing his finger over the open end, he inverted the tube 
in a bath of mercury. The mercury then fell until its upper 
surface was about 30 in. above the level of the mercury in the 
dish, proving the original contention that mercury would 
rise }is as high as water. Torricelli attributed the level of 
the mercury in the tube to the pressure of the air. 

192. Pascal's Experiment. — Pascal argued, if atmospheric 
pressure were the supporting force, that the column of mercury 
would fall somewhat at higher altitudes where the air is rarer. 




Fig. 197. — Torricelli's experi- 
ment. 



1 Galileo Galilei (1564-1642). Born in Italy. Professor of mathe- 
matics at Pisa and Padua. Discovered the laws of falling bodies and the 
pendulum. Constructed the first thermometer. Made many valuable 
discoveries in astronomy. Considered the father of modern physics. 

2 Evangelista Torricelli (1608-1647). Italian physicist and mathe- 
matician. Successor of Galileo. Constructed the first microscope and 
improved the telescope. 



FLUIDS 



183 



Accordingly, he carried Torricelli^s apparatus to the top of 
a tower in Paris and noted that there was a perceptible drop 
in the mercury column. A Torricellian apparatus may be 
used to determine altitude. At places near the sea level, the 
mercury will drop about 1 mm. for every 11 meters of ascent 
or 3-fo of an in. for every 90 ft. of ascent. 

We may verify PascaPs experiment by the fol- 
lowing demonstration. Referring to Fig. 198, 
we have an exhausted tube containing mercury 
to the approximate height of 30 in. or 760 mm., 
the open end dipping into a well of mercury. 
The apparatus is placed upon the plate of an air 
pump and covered with a high bell jar. The 
air is then exhausted within the jar and the 
mercury will fall until it is almost at the level of 
the mercury in the well. If the air is allowed to 
re-enter the bell glass, the mercury will rise to its 
former position. This shows conclusively that 
the rise of liquids in exhausted tubes is due to 
atmospheric pressure. 

193. The Mercury Barometer. — The ordinary 
mercury barometer is simply a Torricellian tube 
mounted as shown in Fig. 199(a). The tube 
terminates at the lower extremity in a generous 
sized reservoir with a flexible bottom (usually 
chamois). At the left, the scale is graduated 
in inches and tenths of inches; at the right, 
the corresponding range in millimeters. The 
protecting cap over the reservoir is constructed 
so as to afford a clear view of the mercury as well 
as the fixed white glass pointer (see Fig. 199(6)) which should 
always coincide with the convex surface of the mercury. 
Before reading the instrument, the thumbscrew must be so 
regulated that the pointer and the surface of the mercury will 
just meet. If the pressure^ of the atmosphere increases, the 



Fig. 19 8.— 
Apparatus t o 
prove that the 
mercury c o 1- 
umn of a Tor- 
ricellian tube is 
supported b y 
atmospheri c 
pressure. 



1 The rear of the mercury reservoir has a porous stopper. 



184 



MECHANICS 



^1^ 



Mr 



Jj- 



mercury will ascend in the tube; if the 
pressure decreases, it will descend in the 
tube. In reading the barometer, great care 
must be exercised that the eye and the con- 
vex surface of the mercury in the tube are 
in a horizontal line. A rising barometer 
indicates fair weather; a falling barometer 
indicates that a storm is approaching. A 
sudden drop at sea indicates a storm of 
unusual severity. The barometer is not a 
sure storm prophet, yet it is very helpful in 
making weather predictions. While it may 
not storm at every low barometer, we know 
that the barometer is low w henever it does 
storm. The mercury barometer is used 
principally for stationary work and where 
the temperature is not sufficient to freeze 
the mercurv. Mercurv freezes at — 39°C. 



■'Poinfer 



MercufL/ 
Reservoir 



Flexible 
'dottom 



Fig. 199(6). — View of mercury reservoir 
in Fig. 199(a). 

194. The Aneroid Barometer. — The aner- 
oid barometer is shown in Fig. 200. It 
consists of a circular box of metal, the sides 
of which are thin, elastic and corrugated. 
^ The air is partially exhausted /rom the 

cury barometer, box and the box is then sealed. One side 




FLUIDS 



185 



is securely fastened to the base and the other side oper- 
ates against a system of levers, etc., which in turn move 
the pointer around the graduated dial. If the outside pressure 
increases, the front face of the box moves in, registering an 
increased pressure on the dial. 
If the outside pressure decreases, 
the face of the box moves outward 
and the pointer moves back due 
to the spring with which it is 
connected. Alieroid barometers 
are graduated by comparison with 
a standard mercury barometer. 
Aneroid barometers are very sen- 
sitive in operation but they fre- 
quently get out of order and need 
re-graduating. They are used for 
work in which it is necessary to 
move from place to place and also in the barograph which 
will be described in the next paragraph. 

195. The Barograph. — The barograph, shown in Fig. 201, 
is an aneroid instrument designed to record automatically 




Fig. 200. — Aneroid barometer. 




Fig. 201. — Barograph or recording barometer. 

the varying pressures of the atmosphere. The drum, revolv- 
ing once every seven days, carries a graduated chart. The 
variations of pressure cause movements of the lever, the free 
end of which carries a pen. Thus the pressure for an entire 
week is self-recorded on the chart. In this barometer the 



186 MECHANICS 

motion is magnified by the use of a series of rarefied 
boxes connected one to another, rather than a single box. 
The barograph may be seen at any government weather 
station. 

196. Standard Conditions of Pressure and Temperature. — 
Since gas volumes are affected both by pressure and tempera- 
ture, it necessarily follows that experimenters doing similar 
quantitative work on gases in different localities would not 
agree in their results, due to different conditions of pressure 
and temperature. In order to obviate this difficulty, standard 
conditions of pressure and temperature have been adopted. 
Uniformity is obtained b}' reducing all numerical w^ork to 
these standards. 

Standard conditions for gases are a pressure of 76 cm. of 
mercury and a temperature of 0°C. or 32°F. For ordinary 
calculations either 76 cm. or 30 in. may be used. 

197. The Magdeburg Hemispheres. — The original Magde- 
burg hemispheres, invented by Otto von Guericke,^ consist of 
two hollow metallic hemispheres fitting tightly together. The 

spheres are 22 in. in diameter and it is 
said that four teams of horses on either 
side were unable to separate them when 
the air was exhausted, due to the pres- 
sure of the air on the outer surfaces. 

The hemispheres shown in Fig. 202 are 
used for ordinary demonstrations and are 
4 in. in diameter. The lower hemisphere 
terminates in a threaded stem which may 
be removed from the base and screwed 
into the plate of an air pump. When the 
'^ hemisphcJros.^ ^"^^ ^^^ ^^ cxhaustcd, the outsidc pressure 

holding them together is so great that 
two students will experience great difficulty in separating 

^ Otto von Guericke (1602-1686). German physicist and astronomer. 
Mayor of Magdeburg;. Made many experiments with liquids and 
gases. Constructed the first air pump in 1605. Inventor of the Magde- 
burg hemispheres which four teams of horses could not separate. 




FLUIDS 187 

them. In computing the total pressure tending to keep 
the hemispheres together, simply assume a circular plane 
whose diameter is the same as the diameter of the hemi- 
spheres. The curved area is not used on account ^of the 
fact that the various pressures, acting at right angles to 
the curved surface, do not all point in the same direction for 
each individual half. The components of these pressures, 
acting perpendicular to the circular plane assumed, comprise 
the total force acting. If the vacuum is perfect, the force 
necessary to pull apart the hemispheres shown in Fig. 202 
will be 4 X 4 X .7854 X 14.7 or 184.72 lb. 

Questions and Problems 

1. State the general characteristics of gases. 

2. What pressure is exerted by the atmosphere at sea level? 

3. What is the average density of air at sea level and at 0°C. ? 

4. What is meant by the height of the homogeneous atmosphere? 

5. Describe an experiment to prove that air has weight. 

6. Describe an experiment to prove that air exerts pressure. 

7. Describe Torricelli's experiment. 

8. Describe Pascal's experiment. 

9. How would you verify Pascal's experiment in the laboratory? 

10. Describe the construction and operation of a mercury barometer. 

11. Describe the construction and operation of an aneroid barometer. 

12. Is the barometer a good weather prophet? Explain. 

13. How would you determine the altitude by means of a barometer? 

14. State clearly what is meant by standard conditions of temperature 
and pressure. Why were standard conditions adopted? 

15. Describe the original Magdeburg hemispheres. How would 3^ou 
compute the force necessary to separate a pair of such hemispheres? 
Assume a vacuum within. 

16. What is the weight of a column of mercury 30 in. high and 1 sq. 
in. in cross-section? What pressure does the column exert? 

17. What is the weight of a column of mercury 76 cm. high and 1 
sq. cm. in cross-section? What pressure does the column exert? 

18. A water barometer reads 33 ft. What is the pressure of the air? 

19. Find the weight of the air in a room 20 X 30 X 12 ft. Assume 
standard conditions. 

20. What will the barometer normally read at the top of the Woolworth 
building 792 ft. high? 

21. If the atmosphere had the same density throughout as at the 



188 



MECHANICS 



A' 



IT 



earth's surface, how high would it extend in miles? Assume the 
height of the barometer to be 30 in. 

22. If the pressure within the Magdeburg hemispheres (Prob. 15) is 
sufficient to sustain a 2-inch column of mercury, what force will be neces- 
sary to separate them? 

198. Boyle's Law.^ — Robert Boyle ^ was the first man to 
make a study of the relation of a confined gas and the pressure 

exerted by the gas. He stated, as 
a result of his investigations, that, 
during all volume and pressure 
changes for a given gas, the product 
of the volume and the pressure will 
remain constant. For example, sup- 
pose that we have 4 cu. ft. of air 
confined under a pressure of 1 almo- 
sphere. The product of the volume 
and pressure will be 4 X 1 or 4. 
Suppose, further, that we increase 
the pressure against the gas to 2 
atmospheres. The volume will be 
reduced, to 2 cu. ft. and the product 
of the volume and pressure will be 
2 X 2 or 4. The above relation is 
usually stated: 

Tine temperature remaining con- 
stant , the volume of a confined gas 
will vary inversely as the pressure 
acting upon it, 

Boyle's law may be stated algebraically: 

V, .y, = P^.P, or FiPi = F2P2. 

The following simple experiment will verify Boyle's law. 
Figure 203(a) represents a glass tube of uniform bore, bent as 
shown and closed at C. Mercury is poured into the open end 

^ Boyles law is not exact, but is close enough for practical purposes. 
2 Robert Boyle (1627-1691). English physicist and chemist. Experi- 
mented extensively in pneumatics. 



XJ 



-5 



u 



T^C 



■B 



Fig. 203. — Apparatus for veri- 
fying Boyle's law. 



FLUIDS 189 

through a funnel and the apparatus adjusted until mercury 
levels A and B are the same. The volume of the confined 
air (proportional to BC) is then under a pressure of 1 atmos- 
phere or 76 cm. of mercury. More mercury is poured into the 
tube, see Fig. 203(6), until the level of the mercury in the 
open part is 76 cm. above the level in the closed part. The 
confined gas is now under a pressure of 2 atmospheres or 152 
cm. of mercury and the volume of the confined air (propor- 
tional to C'B') will be reduced to 3^^ its former size. By 
doubling the pressure we have halved the volume. 

It isr evident, in the above case, that the density of the air 
was doubled when the pressure was doubled. Hence: the 
density of a confined gas varies directly as the pressure upon it. 
Algebraically : 

Di :D2 = Pi :P2. 

If the volume remains constant, the weight of a confined gas 
varies directly as the pressure it exerts. Algebraically: 

W,:W2= Pi: P2. 

Questions and Problems 

1. State Boyle's law. 

2. Describe an experiment to verify Boyle's law. 

3. State the relation between the density of a confined gas and the 
pressure upon it. 

4. State the relation between the weight of a confined gas and the 
pressure exerted by the gas, volume constant. 

5. A vessel contains 5 liters of air at standard pressure. What will 
the volume be if the pressure is changed to (a) 780 mm.? (b) 740 mm.? 

6. 10 cu. ft. of illuminating gas is at a barometer pressure of 30 in. 
What will be the pressure, if the volume is compressed to 7 cu. ft.? 

7. The gasoline tank of an automobile lacks 10 gallons of being full. 
The engine pump communicating with the tank is allowed to run until 
the air gauge on the dash reads 2 lb. Find the weight of the air in the 
tank. 1 cu. ft. of air at standard conditions weighs .0807 lb. 

8. A 10 cu. ft. tank of helium gas (density at standard conditions 
= .0112 Ib./cu. ft.) is at a pressure of 760 mm. of mercury. Helium 
is pumped into the tank until the pressure is 50 lb. Fnd (a) the density 
of the gas; (b) the weight of the gas. 

9. The weight of the air in a room is 50 lb. at 29.5 in. of mercury. 
What will be the weight at 30 in. of mercury? 



190 MECHANICS 

10. A 1922 Buick six cylinder motor has a bore of 3.375 in. and a 
stroke of 4.5 in. Assuming the clearance to be 1 in., what pressure will 
the mixture of air and gasoline vapor exert under full compression? 
What will be the total pressure against the cylinder head? Take the 
atmospheric pressure as 15 lb. Assume constant temperature. 

11. Repeat as above for an eight cylinder Cadillac motor. The bore 
is 3.125 in. and the stroke is 5.125 in. 

199. The Aeroplane. — Since the aeroplane is heavier than 
air, the lifting force is not due to the same reason as in the case 
of a balloon. The lifting force is due to the reaction or thrust 
of the air against the planes as the machine is driven through 



Fig. 204. — Curtiss model 18-r triplane. 

the air at a high rate of speed. The velocity is o])tained by 
means of a special wood propeller driven by a gasoline motor 
of special design. Aeroplanes have proven indispensable in 
war for scouting, boml^ing, etc. Thoy have yet to prove their 
success as heavy cargo carriers. 

The aeroplane shown in the above cut is a Curtiss Model 
18-r Triplane. The over-all length is about 23 ft. and the 
over-all height about 10 ft. The wing span is nearly 32 ft. 
The machine weighs 1,825 lb. and will carry a gross weight of 
2,901 lb. It has a maximum horizontal flight velocity of 
163 m.p.h. and a climbing speed of 15,000 ft. in 10 minutes. 
At 2,500 r.p.m., the motor horsepower is rated at 400. 



FLUIDS 



191 



200. The Balloon. — The lifting ability of a balloon depends 
upon the buoyancy of the air. Just as a body submerged in 
water is buoyed up by the weight of the water displaced, so a 
body submerged in air is buoyed by the weight of the air dis- 
placed. The buoyant force of water is 62.5 lb. for each cubic 
foot displaced. Since air is about 1/773 as heavy as water, 
the buoyant force of a cubic foot of air will be 62.5/773 or 
approximately .08 lb. A balloon containing a vacuum would 
have a tremendous lift, but would be crushed by the outside 




Fig. 205. — U. S. Army type AC dirigible. 

pressure. To obtain a suitable lifting force, the balloon is 
filled with some very light gas as hydrogen, helium, etc. 
Helium is today regarded as the most satisfactory gas. It is 
heavier than hydrogen, but is not inflammable like hydrogen. 
The gas densities given below are at standard conditions, i.e., 
0°C. and 76 cm. of mercury: 

Hydrogen 0056 Ib./cu. ft. 

Helium 0112 Ib./cu. ft. 

Illuminating gas 0500 Ib./cu ft. 

Air 0807 Ib./cu. ft. 

The lift exerted by a balloon depends upon the density of 
the gas with which the balloon is filled. It is equal to the 
weight of the air displaced by the envelope minus the weight 
of the gas in the envelope. The ascensional force is the force 



192 MECHANICS 

with which the balloon leaves the ground. It is equal to the 
lift minus the weight of the envelope, rigging, car, passengers, 
etc. 

Figure 205 shows a type AC dirigible recently built for the 
U. S. Army. It is 167 ft. long, has a gas capacity of 187,000 
cu. ft. and a cruising speed of about 65 m.p.h. The car is 
entirely closed and will carry 8 people. 

An Army observation balloon of the R t^^pe is 90 ft. long, 29 ft. in 
diameter and has a gas capacity of 37,000 cu. ft. The envelope, rigging, 
car, etc., weigh 1,000 lb. 

Suppose we wish to find the total lifting effect and the ascensional force 
at ground conditions, if the envelope is filled with pure hA^drogen. 

Solution: 

.0807 - .0056 = .0751 lb. lifting effect per cu. ft. of hydrogen, 

37,000 X .0751 =2,778.7 lb. total lifting effect, 

2,778.7 - 1,000 = 1,778.71b. ascensional force. 

Suppose now that we wish to find the ascensional force of the above 
balloon if filled with commercial hydrogen (98 per cent. pure). Assume 
the 2 per cent, impurity in the h3^drogen to be air. 

Solution: 

37,000 X .98 = 36,260 cu. ft. of pure hydrogen, 

36,260 X .0751 = 2,723.1 lb. total lifting effect, 

2,723.1 - 1,000 = 1,723.1 lb. ascensional force. 

Questions and Problems 

1. Explain the lifting effect of an aeroplane. 

2. What is meant by the lift and ascensional force of a ballon? 

3. State how each of the above is determined. 

4. State the various gases used in ballons. What is the advantage 
of each? 

6. Why does a submarine either sink to the bottom or float on the 
surface when the power is cut off, while a balloon, on ascending, will 
reach a certain height and float? 

6. A spherical ballon 50 ft. in diameter is filled with pure hydrogen. 
Find: (a) the buoyant force of the air; (6) the weight of the hydrogen; 
(c) the total lift. 

7. Repeat Problem 6, using pure helium. 

8. Repeat Problem 6, using 98 per cent, pure hydrogen. 

9. Repeat Problem 6, using 92 per cent, pure helium. 

10. A spherical balloon has a capacity of 40,000 cu. ft. and is filled with 
commercial hydrogen (98 per cent. pure). The balloon and car weigh 



FLUIDS 



193 



450 lb. Two passengers weighing 160 lb. each are carried. Find: (a) 
the total lift; (b) the ascensional force. 

11. An 84,000 cu. ft. U. S. Army dirigible is filled with pure hydrogen. 
The gas bag, car, motors, crew, etc., weigh 6,300 lb. What is the 
ascensional force? 




F 



Wafer 



'Wafer 



A 5 

Fig. 206. — Siphon. 



201. The Siphon. — The siphon is a device used to transfer 
liquids from one receptacle to another. It consists of a bent 
tube of unequal arms. A rubber tube may be used for 
ordinary liquids. Referring to Fig. 
206, let us study the operation of 
a siphon. Assume that water is 
flowing from vessel A into vessel 
B through the bent tube. Since 
EF is shorter than CG, it is evident 
that the weight of the liquid in the 
latter arm is greater than in the 
former. The upward pressure at F 
is the atmospheric pressure minus 
the pressure of the column EF . The 
upward pressure at G is the atmos- 
pheric pressure minus the pressure 
of the column CG, Therefore the 

upward pressure at F exceeds the upward pressure at G. 
It is evident that there will be a flow from the point of 
greater pressure at F to the point of less pressure at G. 
The flow will continue until F and G are at the same level and 
hence at the same pressure. If EF is higher than 34 ft., the 
siphon will not work, as water will not rise above that point 
in an exhausted tube. If mercury is used, EF must not 
exceed 30 in. 

202. The Air Pump. — The so-called air pump is an appa- 
ratus for exhausting air from vessels. It was invented in 
1650 by Otto von Guericke. With a good pump, a very high 
degree of vacuum may be obtained. Figure 207 shows the 
construction of a simple air pump. It consists of a cylinder 
(C) with a tight-fitting piston (P). Fi and F2 are valves. 
On the upward stroke, a vacuum is formed under the piston 

13 



194 



MECHANICS 



in the cylinder, T^i closes automatically and T"2 opens, and the 
air in the bell glass (B) passes in part to the cylinder to fill up 
the vacuum and equalize the pressures. On the downward 
stroke, Vo closes and T\ opens, allowing the air in the cylinder 
to pass out. The above process is repeated, each time a 




Fig. 207. — Air pump. 

portion of the air in B being removed. Since only a fraction 
of the air is removed at each stroke, it will be seen that a 
perfect vacuum can never be obtained. The air pump may 
be converted into a compression pump by opening 1% to the 
atmosphere and attaching the receptacle, into which the air is 
to be compressed, to the tube at Fi. 

203. The Air Brake. — Figure 208 shows the construction of a 

simple Westinghouse air 
brake. T is a tank car- 
ried by each separate unit, 
in which the engine pump 
maintains, through pipe 
(P), a pressure of about 
75 lb. While the engine 
pressure is acting in P, 
TT ono w *• u • u 1 the triple valve V main- 

FiG. 208. — vV estinghouse air brake. ' 

tains communication 
between P and T, If the pressure is cut off in P, either 
at the engine or by accidental uncoupling of the pipe line, 




FLUIDS 



195 



communication between P and T is cut off at V and communi- 
cation is opened between the tank {T) and the cyUnder (C). 
Thus the compressed air in T forces the piston to the left in 
the cyhnder, setting the brake shoes against the wheels. To 
remove the friction of the brakes, pressure is again turned on 
in P. The connection between T and C is thus closed, the air 
from C exhausts at E and the spring {S) moves the piston 
back to its original position. 
204. Liquid Pumps with Reciprocating Parts. — Recipro- 




Water 
Fig. 209.— Lift pump. 




Fig. 210. 



Wa-ter 
-Force pump. 



eating pumps ordinarily depend upon atmospheric pressure 
for their operation. The lift pump (Fig. 209) consists of a 
cylinder with a tight-fitting piston. The piston carries a 
valve (7) and the lower part of the cylinder has a valve (Fi). 
Whenever the piston is drawn up, V closes and Vi opens, water 
being forced up into the cylinder by the pressure of the air. 
On the down stroke Vi closes, trapping the water already in the 
cyhnder. At the same time V opens, allowing the trapped 
water to pass into the upper part of the cylinder. On the 
next up stroke V closes, trapping the water above the piston 
and forcing it out for delivery. Simultaneously Fi opens and 
the previous cycle is repeated. The pump will not operate if 



196 



MECHANICS 



Vi is much more than 30 ft. above the surface of the supply 
water. 

Figure 210 shows a force pump. Vi opens as the piston is 
drawn up, and water is forced up into the cyHnder by atmos- 
pheric pressure; at the same time V closes. On the down 
stroke Vi closes, trapping the water in the cylinder, and V 
opens, allowing the water to pass into the delivery tube. 




Fig. 211. — Goulds "Triplex" water pump. 

On the next up stroke V closes, Fi opens and the cycle is 
repeated. The air chamber at the top of the delivery tube 
causes the water flow from the orifice in a steady stream^ 
instead of intermittently. There is no reasonable limit to the 
height that the water may be forced, provided that Vi is 
not too far above the water supply. 

Figure 211 is from a photograph of a Goulds single-acting 
^^ Triplex'^ pump. This type of pump is very widely used. 
It has a maximum working pressure of 300 lb. At this pres- 
sure, it will elevate water to a height of nearly 700 ft. Atten- 
tion is called to the large air chamber in front. 



FLUIDS 



197 



205. Liquid Pumps with Rotating Parts. — Pumps with 
rotating parts have come into great prominence during the 
past few years. Figure 212 shows a gear pump. Its opera- 
tion will be clear from an examination of the figure. This type 




^'quic/Ouf 




Fig. 212. — Gear pump. 



Fig. 213 — Exposed section 
of a centrifugal pump. 




Fig. 214. — Worthington centrifugal fire pump. 



of pump is used when the volume of liquid to be pumped is 
small. Many automobiles make use of gear pumps to actuate 
the oil circulation in the motor. 

Centrifugal pumps have been used for some time for low 
pressure purposes, such as circulating the water in internal 
combustion engines. They are now accepted as the proper 
kind of pumps for delivering liquids against high heads. 



198 



MECHANICS 



Those designed for high pressure are either electrically or 
steam turbine driven. In principle the centrifugal pump is a 
reaction turbine working backwards. Its construction will 
be seen from Fig. 213. The liquid enters at the center (shown 
by the dotted circle). As it strikes the blades, it is thrown 
toward the circumference of the casing. The liquid passes 
into the delivery pipe (D) with a pressure depending upon the 
design and speed of the pump. 

Figure 214 shows a Worthington centrifugal pump used for 
fire purposes. It is turbine driven and will furnish 1,500 
gallons of water per minute against a pressure of 100 lb. 

206. Instruments for Measuring Pressure. — The commer- 
cial form of pressure gauge used on steam boilers, gas 





Fig. 215. — Pressure gauge. 



Fig. 216. — Vacuum gauge. 



tanks, etc. (Fig. 215) is graduated in Ib./sq. in. This type of 
gauge is designed to register the pressure in excess of atmos- 
pheric. Thus, if the gauge reads 100 Ib./sq. in., the actual 
pressure of the gas under test is 115 Ib./sq. in. The vacuum 
gauge (Fig. 216) is used to measure the degree of rarefaction 
in a closed gas container and is graduated in inches. Zer(] 
in. corresponds to a pressure of 15 lb. and 30 in. corre- 
sponds to a vacuum or zero lb. Hence, if the gauge reads 
15 in., the actual pressure is 15/30 of 15 or 7.5 lb. 

Figure 217 shows a laboratory model to illustrate the pres- 
sure and vacuum gauge. The hollow curved tube communi- 
cates with the gas under test. Whenever there is an increase 



FLUIDS 



199 




Fig. 217.— Work- 
ing parts of a pres- 
sure gauge. 



in pressure, the tube tends to straighten, as a piece of garden 
hose when the water is turned on. The free end of the tube 
moves out and communicates its motion to a lever, the teeth 
on the extremity of which mesh with a 
pinion carrying the indicator. If the pres- 
sure decreases, the tube will bend in, allow- 
ing the indicator to move in the opposite 
direction, due to the spirally wound spring 
with which the indicator is connected. 
The indicator will remain stationary when 
a vacuum has been obtained. 

The open-arm manometer shown in Fig. 
218, may be used for pressures above or 
below atmospheric. It consists of a tube 
bent as shown, open at A, containing a 
liquid of known density and connected to 
the gas container at C. If the pressure in 
the container is greater than atmospheric, the liquid in 
arm A will rise above the level of the liquid in arm B. 

The height of the liquid in A over the 
height in B determines the excess of the 
pressure in the container over atmos- 
pheric. The atmospheric pressure plus 
the excess pressure indicated gives the 
actual pressure exerted by the gas. If 
the confined gas is under a pressure of 

Uless than one atmosphere, the liquid in B 
will rise above the level in A . The excess 
of the atmospheric over the actual pres- 
sure exerted by the gas under test is com- 
puted from the difference in height of 
the liquid columns. The actual pressure 
exerted will be the atmospheric pressure, 
as determined by the barometer, minus 
the pressure as indicated by the liquid columns. 

The closed-arm manometer shown in Fig. 219 is used to 
measure higher pressures than the open-arm manometer. 



Fig. 218. — Open-arm 
manometer. 



200 



MECHANICS 



B 



The tube is connected at C, as in the previous case, but arm A 
is closed. The Hquid used is mercury and the space above 

the mercury in the closed arm is filled 
with air. The pressure is determined 
by means of the change in the height of 
the mercury columns and by the change 
in the air volume. At ordinary atmos- 
pheric pressure the mercury levels will be 
Uthe same. Before C is connected to the 
confined gas, the position of the mercury 
levels is noted as well as the volume 
of the air in arm A. The barometer 
reading is also taken. As the arm is uni- 
form in bore, the volume is proportional 
to the length of the air column. Length 
Fig. 219.— Closed-arm may therefore be used instead of volume, 
manometer. .^j^^ pressure is Computed from the 

change in the mercury levels and from the change in the 
length of the air column according to Boyle^s law. 



Questions and ProbJems 

1. Explain the action of the siphon. 

2. Describe the construction and operation of an air pump. How 
may the air pump be converted into a compression pump? 

3. Why is it impossible to obtain a perfect vacuum with an air pump? 

4. Describe the construction and operation of the Westinghouse air 
brake. 

5. Into what two classes may liquid pumps be divided? Give the 
construction and operation of the liquid pumps described in this chapter. 

6. What is the difference between a pressure gauge and a vacuum 
gauge? 

7. Describe the principle upon which the commercial gauge works. 

8. Describe how pressure is determined with an open-arm manometer. 

9. Describe how pressure is determined with a closed-arm manometer. 

10. If in Fig. 206 EF is 20 in. and CG is 40 in., compute the upward 
pressure at F and the upward pressure at G. Barometer reads 30 in. 

11. Assuming the diameter of C in Fig. 208 to be 12 in. and the pres- 
sure in T to be 75 Ib./sq. in., what is the total pressure against the 
piston, if connection between T and C is open? State two reasons why 
the piston rod will not transmit the entire amount of the total pressure 



FLUIDS 201 

12. If the reading of a pressure gauge is 25 Ib./sq. in., what is the total 
pressure exerted by the gas under test? 

13. If a vacuum gauge reads 20 in., what is the actual pressure of the 
gas under test? 

14. The open-arm manometer (Fig. 218) is connected at C to a gas 
tank through a two-way valve. When the valve is opened to the air, 
the water levels are the same. When the valve is opened to the gas 
tank, the level in A is 3 in. above the level in B. Find the actual gas 
pressure in the tank. 

15. Repeat Problem 14, if the level in B is 2 in. higher than in A. 

16. Find the pressure of the confined gas in problem 14, if the level in 
A is 20 mm. higher than the level in B, Give answer in Kg./cm^. 

17. Before connection to a gas tank, the level of mercury in the closed 
arm of a manometer containing mercury is 2 cm. higher than in the 
opposite arm. After connection to the gas tank, the difference in mercury 
levels is 10 cm. and the air column is only half as long What is the actual 
pressure of the gas in the tank? 



CHAPTER XXII 

FALLING BODIES; CENTRIFUGAL FORCE; THE 

PENDULUM 



SECTION 1. FALLING BODIES 

207. Effect of Gravity on Falling Bodies. — Peculiar as it may 
seem to the student, gravity acts with equal effect on all 
bodies. If a heavy body and a light body are dropped from 
the same height at the same time, they will reach the ground 
simultaneously, provided that we neglect the 
resistance of the air. The air exerts a retard- 
ing effect on falling bodies, which depends 
upon the surface exposed. For example, a 
piece of aluminum weighing 5 lb. will fall 
somewhat more slowly than a piece of gold 
weighing 5 lb., due to the fact that the. alumi- 
num offers more area to the air. In a vac- 
uum, however, they will fall with identical 
speeds. This may be proved as follows. 
Figure 220 represents a glass tube 4 ft. long 
and closed at one end. The lower end termi- 
nates in a petcock with an air pump attach- 
ment. Within the tube is a feather and a 
penny. If the tube is quickly inverted, the 
raUiT'to^^h^^^hat P^^ny will be seen to fall more rapidly than 
all bodies fall at the feather. If the air is then withdrawn 
vacumn^^ ^^^^ *" ^ from the tube and the experiment repeated, 
the feather and penny will be seen to fall 
with equal velocities. Hence we may arrive at the conclusion 
that bodies, falling from the same height in a vacuum, icill fall 
equal distances in equal lengths of time. As air resistance is 
difficult to determine, it will not be taken into account in 
the solution of problems. 

202 




FALLING BODIES 



203 



208. The Acceleration of Gravity. — It has been determined 
by experiment that the acceleration of gravity is constant; 
that is, a freely falling body will be imparted an equal gain in 
velocity for each succeeding second. Starting with a velocity 
of zero, the body will have a velocity of 32.16 ft. /sec. at the 
end of the first second; 64.32 ft. /sec. at the end of the second 
second, etc. Thus we have an example of uniformly acceler- 
ated motion. The force of gravity varies slightly from place 
to place. At New York, the acceleration of gravity 
(symbol "g'^) is about 32.16 ft. per second per second or 980 
cm. per second per second. It is usually written 32.16 ft./ 
sec;2 or 980 cm. /sec. ^ During the first second, a freely falling 
body will travel 16.08 ft. or 490 cm. 

209. Freely Falling Bodies. — We may now construct the 
following table for freely falling bodies, allowing D to repre- 
sent ^"2 Q 01* the distance passed over during the first second. 
Since the acceleration is constant, the body will gain 2D 
in velocity each second and will pass over during each succeed- 
ing second 2D more than in the previous second. 



t 


V 


s 


S 


in. which 


1 

2 


2D 
4D 


D 
3D 


D 
4D 


3 


6Z) 


5D 


9D 




4 


8D 


ID 


16D 





t = the time in seconds, 

V = the velocity at the end of any 

particular second, 
s = the distance passed over dur- 
ing any particular second, 
S = the total distance passed over 
at the end of a given number 
of seconds. 



Substituting }^ g for D, it is evident that: 

1. V = gt 

2. 8 = y2g{2t - 1) 

3. S = y^ gt\ 

210. Bodies Rolling Down an Incline. — The acceleration 
for bodies rolling down an incline will be less than g. It is 
found by multiplying g by the height iji) of the plane over the 
length {I) of the plane. 

Acceleration on an incline = g X h/l 



204 MECHANICS 

Since h/l is equal to the sine of the angle of inclination, the 
following formula ma}" be used : 

Acceleration on an incline = g sin 6 

211. Bodies Projected Vertically Downward. — In case a 

body is projected vertically downward from a state of rest 
{e.g., a ball thrown down from a high building), the formulas 
given in the last paragraph must be changed to read: 

1. V = Initial velocity + gt 

2. s = Initial velocity + Mg(2^ - 1) 

3. aS = Initial velocity X ^ + M^^.^ 

212. Bodies Projected Vertically Upward. — If a bod}- is 
projected vertically upward, the acceleration of gravity^ is 
negative; i.e., the body will lose 32.16 ft. /sec. or 980 cm. /sec. 
until a velocity of zero is obtained. Starting from zero the 
body will return, the time of ascent and descent being equal, 
as well as the initial and final velocity. The time of ascent 

Questions and Problems 

(Assume ^ = 32 ft./sec.^ or 980 cm./sec.^) 

1. Describe an experiment to show that all bodies fall at the same 
rate in a vacuum. 

2. Explain what is meant by the acceleration of gravity. 

3. Find v and S for a body freely falling from rest for 6 sec. 

4. A bod}' is rolling down a frictionless inclined plane 20 ft. long and 
5 ft. high. Give the acceleration in feet and centimeters. 

6. A ball is dropped from the Wool worth Building at a point 600 ft. 
above the sidewalk. Find the time of descent and the velocity on strik- 
ing the walk. 

6. In problem 5, how far did the ball travel in the fourth second? 
In the fourth and fifth seconds together? 

7. How long will it take a spherical bodj' to roll do\NTi a 10 per cent, 
grade, if the grade is 100 meters long? 

8. What distance did the body in the previous problem pass over 
during the second second? What was the final velocity? 

9. A stone is thrown vertically downward from a height of 200 ft. 
with an initial velocity of 10 ft. /sec. (a) With what velocity did it 
strike the ground? (6) How long was it in the air? (c) How far did it 
travel in the second second? 



FALLING BODIES 



205 



10. A body is projected vertically upward with a velocity of 100 
meters per sec. (a) How long was it in rising? (b) How high did it 
rise? (c) How long was it in the air? (d) How far did it travel in the 
fifth sec? (e) With that velocity did it strike the ground. 

11. A golf ball is dropped from a height of 200 ft. and }4 sec. later a 
second ball is dropped. How far apart will they be when the first ball 
strikes the ground? 

213. Bodies Projected Horizontally. — If a body is dropped 
from a certain point and another body is projected horizontally 
from the same height at the same time, both bodies will reach 




Fig. 221. — AEFC represents the path of a body projected horizontally from A, 



the ground at the same instant. The second body will advance 
horizontally at a uniform rate of speed, but will curve down- 
ward in a parabolic path due to the action of gravity. 

AEFC (Fig. 221) represents the path of a body projected 
horizontally from A with a velocity of 75 ft. /sec. For con- 
venience we will assume that A is 144.72 ft. above the earth, 
so that the time in the air will be exactly three seconds. At 
the end of the first sec, the body will have advanced horizon- 
ally a distance of 75 ft. and gravity will have drawn it 16.08 
ft. toward the earth. Therefore, at the end of the first 
second, the body will be at E. At the end of the second 
second the body will have advanced 75 ft. more horizontally 
and gravity will have drawn it 48.24 ft. nearer the earth or 
64.32 ft. below the line AD, Hence, at the end of the second 



206 



MECHANICS 



second, the body will be at F. Similarly, it can be shown that 
the body will be at C at the end of the third second. The 
distance BC is called the range. The actual path of the body 
is called the trajectory. 

Suppose we have a body projected horizontally from a height of 
257.28 ft with a velocity of 1,000 ft./sec. It is required to find: (1) 
the time of descent; (2) the vertical velocity on striking; (3) the range. 

Solution : 

I. S = 1/2 gt,^ 257.28 = 16.08 ^2, t'^ = IQ, t = 4: sec. 

2.v = at,v = 32.16 X 4 =128.64 ft./sec. 

3. 1,000 X 4 = 4,000 ft. 

214. Bodies Projected so that the Angle of Elevation is 
Less than 90°. — Suppose a body, projected with a velocity 




Fig. 222. — ARK represents the path of a body projected from A at an angle 

of 40° with the horizontal. 



of 200 ft./sec. from A, makes an angle of 40° with the ground 
(Fig. 222). Were it not for gravity, the body would continue 
at the same rate indefinitely in the straight line AG. On 
account of gravity, the body will be 10.08 ft. below B at the 
end of the first sec, 04.32 ft. below C at the end of the second 
second; 144.72 ft. below D at the end of the third sec, etc. 
The actual path of the body will be parabolical. 



FALLING BODIES 207 

To solve problems in the above case, it is necessary to resolve the 
initial velocity into vertical and horizontal componepts (F and H) and 
consider them as separate velocities. Using data as in the previous 
paragraph, suppose we wish to find: (1) the time of ascent; (2) the time 
in the air; (3) the final vertical velocity; (4) the greatest vertical height 
{h)] and (5) the range. 

Solution: 

First it is necessary to resolve 200 ft. /sec. into its vertical and hori- 
zontal components. By trigonometry: 

F = 200 X cos 50°, F = 200 X .643 = 128.60 ft./sec. 
^ = 200 X cos 40°, F = 200 X .766 = 153.20 ft./sec. 

We may now consider the body to have an initial vertical velocity of 
128.60 ft./sec. and a uniform horizontal velocity of 153.20 ft./sec. 

1. t = v/g, t = 128.60/32.16 = 4 sec. 

2. 4 X 2 = 8 sec. 

3. Final vertical velocity = initial vertical velocity = 128.60 ft./sec. 

4. ,Sf = K ^^,' S = 16.08 X 16 = 257.28 ft. 

5. 153.2 X 8 = 1,225.60 ft. 

Problems 

(Assume g = S2 ft. /sec. ^ or 980 cm. /sec. 2) 

1. Construct a graph showing the path of a body projected horizontally 
from a point 400 ft. above the ground with a velocity of 100 ft./sec. 

2. Construct a graph showing the trajectory of a body projected at an 
angle of 50° with the ground with a velocity of 400 ft./sec. 

3. A projectile is discharged with a horizontal muzzle velocity 100 
meters /sec. at a height of 500 meters above the ground. Find: (a) the 
time of descent; (6) the vertical velocity on striking; (c) the random or 
range. 

4. A bullet discharged horizontally with a velocity of 800 ft./sec. 
has a range of 4,000 ft. Find: (a) the time the bullet is in the air; (b) 
the height of the bullet before discharge. 

6. A ball is shot from a cannon at an angle of elevation of 60° with an 
initial velocity of 1,200 meters /sec. Find: (a) the time of ascent; (6) 
the time in the air; (c) the final vertical velocity; (d) greatest vertical 
height reached; (e) the range. 

6. A baseball is batted from the home plate to the center fielder, a 
distance of 200 ft., with an average horizontal velocity of 60 ft./sec. 
Assuming that the ball starts 4 ft. above the ground and is caught 4 ft. 
above the ground, what is the greatest vertical height reached by the 
ball? 




208 MECHANICS 

SECTION II. CENTRIFUGAL FORCE 

215. Nature of Centrifugal Force. — The particles composing 
a rotating boch^ have a tendency to ''% off'' in a straight 
line at a tangent to the circle described by the individual 
particle. Suppose, for example, that we take particle P on 

the circumference of a rotating fly 
wheel, as shown in Fig. 223. Since 
every body tends to continue its 
motion in a straight line (Newton's 
first law of motion), it is evident that 
P has a tendency to leave the rim of 
the wheel and travel in a line shown 
by the arrow A. This tendency, 
which is reallv a reaction due to 

a fe'„°deSTo''-4t':/at'^: ^'^^' g^ves rise to a pull a^cay from 
tangent" due to centrifugal the Center of the circle and must be 
^^^^^- overcome by an equal and opposite 

force toward the center. The force with which the particle 
tends to leave the center s known as centrifugal force. The 
equal and opposite balancing force is known as centripetal force, 

216. Effects and Uses of Centrifugal Force. — Centrifugal 
force occasions many interesting results, some of which are 
made use of mechanically. Centrifugal force no doubt caused 
the earth, when it was in a plastic state, to become bulged at 
the equator. On curves, railway tracks are constructed 
higher on the outside than on the inside so as to act against 
the overturning tendency of centrifugal force. The curves at 
the Sheepshead Bay automobile track were built at a slant to 
prevent skidding, as racing cars travel at a speed in excess of 
100 m.p.h. 

Laundries make use of centrifugal force in drying clothes. 
The wash is placed in a largo perforated cylinder. As the 
cyhnder rotates, the water tends to fly out. The cream sepa- 
rator operates on the principle of centrifugal force. The 
milk is rotated at a high rate of speed. The heavier portion 
gathers at the edge of the container, leaving the lighter cream 
at the center where it is drawn off. 



CENTRIFUGAL FORCE 



209 



Figure 224 shows a model of a governor used on low-speed 
steam engines to maintain a constant fly wheel speed. Due 
to centrifugal force, the balls tend to move out as the speed 
of rotat'on increases. The movement of the balls operates a 
lever which in turn operates the steam valve, cutting down the 
amount of steam admitted to the cylinder. If the speed 
becomes too low, the balls return toward the center of rota- 
tion, admitting a greater quantity of steam. 

Medium- and high-speed engines make use of a shaft 




Fig. 224. — Model of a steam governor. 

governor carried by the flywheel. It operates on the eccentric 
in such a way that the steam cut-off is retarded or accelerated 
according to conditions. If the engine tends to speed up, the 
governor, acting centrifugally, cuts off the steam sooner; 
if the engine tends to slow down, the action is reversed. 
Early cut-off decreases the /i.p. and a later cut-off increases 
the h.p. 

Many automobile trucks have a centrifugal arrangement to 
prevent excessive speed. A governor is used which limits 
the supply of air and gas delivered to the cylinders. Certain 
automobiles have also a centrifugal device for regulating the 
advance and retard of the spark. 

14 



210 MECHANICS 

217. Determination of Centrifugal Force. — In actual prac- 
tice, we are usually interested in bodies revolving about some 
fixed point. Centrifugal force varies directly as the weight of 
the body and as the square of the velocity of revolution; 
inversely as the radius of revolution. Since centrifugal 
force is equal to centripetal force, one formula will suffice for 
both. 

C entr if uqal force = , 

in which, W = the weight of the body; v = the linear velocity 
of the body at its center of mass; g = the acceleration of 
gravity; and r = the radius of revolution (the distance from 
the center of mass to the center of revolution). In the English 
system, pounds and feet are used; in the c.g.s. system, grams 
and centimeters are used. 

Questions and Problems 

(Assume </ = 32 ft. /sec- or 980 cm. /sec.-) 

1. What is centrifugal force? Carefully explain the cause of centri- 
fugal force. 

2. What is centripetal force? How does it compare in value with 
centrifugal force? Is centripetal force useful? Explain. 

3. Give several instances showing the practical application of centri- 
fugal force to mechanisms. 

4. Give the formula for centrifugal force. 

6. A body weighing 10 lb. is whirled around at the rate of 100 r.p.m. 
If the radius of revolution is 5 ft., what centrifugal force is developed? 

6. An automobile weighing 3,000 lb. is rounding a curve at the rate of 
40 m.p.h. If the radius of curvature is J^ mile, what friction must there 
be between the road and tires to prevent skidding? 

7. A flywheel weighing 300 lb. is rotating 1,500 times per minute. 
There is an unbalanced weight of 5 lb. acting 3 ft. from the center. 
Find the load on bearings due to centrifugal force. 

SECTION III. THE PENDULUM 

218. The Pendulum. — The ordinary pendulum consists of a 
bob suspended by a thread, cord, rod, etc., and capable of 
oscillation about a fixed point. If the weight of the sup- 
porting medium is negligil)le, the length of a pendulum is the 



THE PENDULUM 



211 



distance from the point of support to the center of mass of the 
bob. The amplitude is the hnear distance from the point of 
rest to the farthest position of swing. The "period or time 
of vibration is the number of seconds necessary for a complete 
swing over and back. A ^^ seconds pendulum^' is one which 
makes a half -vibration in one second. Thus, the period of 
a ^^ seconds pendulum'' is 2 seconds. The length of such a 
pendulum at New York is approximately 39 in. or 100 cm. 
219. Why a Pendulum Vibrates. — A vibrating pendulum 




/ D 



Fig. 225. — Diagram to show how gravity produces the oscillations of a 

pendulum. 

is a good example of what is known as simple harmonic motion. 
Harmonic motion is a forward and back motion, in which 
the maximum velocity is at the center, decreasing to zero at 
either end. 

A pendulum set into vibration will continue to vibrate 
with succeedingly smaller amplitudes, until it comes to a posi- 
tion of rest. The force tending to continue the vibration is 
gravity and the force eventually bringing the pendulum to 
rest is friction. 

Let us see exactly how gravity acts with respect to a pendu- 
lum. Suppose (Fig. 225) that the pendulum bob is at rest at 



212 



MECHAXICS 



B. Gravity (G), acting vertically downward, produces 
merely a tension in the supporting cord OB. and there is no 
tendenc}' toward motion. If now the bob is drawn over to A, 
gravity may be resolved into two components, AD' and AE', 
AD' produces tension in OA, while AE'^ acting tangent to tl 
arc AB, tends to produce motion. As A swings to the right, 
AE' decreases in value, until at B it is zero. At B there is 
sufficient kinetic energy- (friction neglected) to carry the bob 
to C. At C gravity may again be resolved into two com- 
ponents, CD and CE. CD produces tension in OC^ while 
CE tends to produce motion. As C swings to the left, CE 
decreases in value, until at B it is zero. At B the bob has 
sufficient kinetic energy (friction neglected) to carr\^ it to .4, 

whereupon the cj'cle is begun again. 
Were it not for friction, the to-and-fro 
motion would continue indefinitely. In 
actual practice, the arc of oscillation 
gradually decreases on account of fric- 
tion, eventually bringing the pendulum 
to rest. The pendulum of a clock and 
the balance wheel of a watch receive 
energy from an outside source (spring) 
and continue to vibrate until the energ>' 
of the spring is exhausted. 

220. Laws of the Pendulum. — The 
mass of a pendulum does not affect its 
vibration rate. This is proved by the 
following simple experiment. Referring 
to Fig. 226, we have two pendulums of 
equal length but different weight. They 
for studying laws of the are oscillated through small amplitudes 
pendulum. ^^^ ^y^^ vibrations of each are counted for 

one minute. It will be found that each pendulum has made 
the same number of vibrations. 

For small amplitudes, the vibration rate is independent of 
the arc of swing. This may be proved by oscillating one of the 
pendulums in Fig. 215 thi*ough different short arcs. The 



'5 




Fig. 226. — Apparatus 



THE PENDULUM 



213 



number of vibrations per minute will be the same in each 
case. 

We know from experience, however, that the vibration rate 
of a pendulum depends upon the length 
of the pendulum: the shorter the length 
the greater number of vibrations in a 
given time. Suppose (Fig. 227) we have 
three pendulums. A, B and C, of like 
weight and material and with lengths of 
81, 64 and 49 cm. respectively. The square 
roots of the lengths are then in the ratio 
of 9 : 8 : 7. Each pendulum in turn is set 
swinging through a small amplitude and 
the number of vibrations per minute 
counted. The periods of vibration are 
then determined. It will be found that 
the periods are in the ratio of 9:8:7, 
proving that the period of vibration is 
directly proportional to the square root of 

the length. Fig. 227.— Appa- 

Since the force of gravity varies from w^'oUhrpenS^^ 
place to place, it will be evident from Fig. 
225 that an increase in the value of ^^g'' will result in a 
decrease in the period of vibration. The period varies 
inversely as the square root of the acceleration of gravity. 



B 




Laws of The Pendulum Summarized 

1 . The period of vibration is independent of the mass, 

2. For small amplitudes, the period of vibration is independent 
of the amplitude, 

3. The period of vibration varies directly as the square root 
of the length, 

4. The period of vibration varies inversely as the square root 
of the acceleration of gravity. 

The laws in 3 and 4 above may be incorporated in the 
following equation: 



T . 2,^, 



214 MECHANICS 

in which T is the period or time of a complete vibration in 
seconds; I the length of the pendulum; and g the acceleration 
due to gravity at the particular locality. Thus, knowing the 
length and period of a pendulum, the value of g maj^ be 
readily determined. 

Questions and Problems 

(Assume ^ = 32 ft./sec.^ or 98° cm./sec.^) 

1. What is a pendulum? 

2. Define the following terms with respect to a pendulum: length, 
ampHtude, period of vibration, seconds pendulum. 

3. Show that a pendulum illustrates simple harmonic motion. 

4. What is the length of a seconds pendulum at New York? 

5. State the laws of the pendulum. 

6. Incorporate the laws of the pendulum in a single equation. 

7. Two pendulums are respectively 4 and 9 in. in length. Compare 
their periods of vibration. 

8. The period of a pendulum is 2 sec. What will the period be if 
the length is doubled? 

9. A pendulum makes 10 v. p.m. How many v.p.m. will it make if 
the length is made one half as great? 

10. Find the length of a second pendulum if g = 980 cm.; 990 cm. 

11. A suspended plumb line makes 10 complete vibrations in one 
minute. How long is it? Give answer in ft. 

12. A plumb line 40 ft. long is dropped from a chimney. What is 
its period of vibration? How many v.p.m. does it make? 

13. A clock pendulum is 12 in. long. How many seconds will the 
clock gain in 24 hr., if the length is shortened to 11 in.? 



APPENDIX 

USEFUL INFORMATION 

Pi, TT = 3.1416. 

Circumference of a circle = rf X tt. 

Area of a circle = xr^ or d^ X .7854. 

Area of a sphere = 4xr2. 

Volume of a sphere = 4/37rr3. 

Lateral surface of a cylinder = circumference of base X altitude. 

Volume of a cylinder = area of base X altitude. 

ENGLISH AND METRIC EQUIVALENTS 

1 inch =2.54 cm. 1 meter = 39.37 in. 

1 mile = 1.61 Km. 1 kilometer = .62 mi. 

1 pound = 453.6 g. 1 kilogram = 2.20 lb. 

1 liquid quart = .946 1. 1 liter = 1.057 liquid qt. 

DENSITY OF COMMON SUBSTANCES 

1 cu. ft. of water at 4°C. weighs 62.4 lb. 

1 cu. in. of water weighs .036 lb. 

1 c.c. of water at 4°C. weighs 1 g. 

1 cu. ft. of air at standard conditions weighs .0817 lb. 

1 c.c. of air at standard conditions weighs .00129 g. 

1 gallon (231 cu. in.) of water weighs 8.32 lb. 

1 cu. ft. of brass (cast) weighs 527 lb. 

1 cu. ft. of copper (cast) weighs 553 lb. 

1 cu. ft. of ice weighs 57.2 lb. 

1 cu. ft. of iron (cast, gray) weighs 442 lb. 

1 cu. ft. of mercury weighs 848 lb. 

1 cu. in. of mercury weighs .49 lb. 

UNITS FREQUENTLY USED 



it 



g'' at New York = 32.16 ft./sec.^ 

g'' at New York =980.2 cm./sec.2 
Length of a seconds pendulum at New York =99.3 cm. 
1 dyne = .00102 g. 

215 



216 MECHANICS 

1 gram =. 980 dynes. 

1 pound = 445,000 dj-nes. 

1 erg = 1 d\Tie centimeter. 

1 joule = 10,000,000 ergs. 

1 horsepower = 33,000 ft. Ib./min. 

1 horsepower = 550 ft. lb. /sec. 

1 horsepower = 746 watts. 

1 kilowatt = 1,000 watts. 

1 horsepower = 746/1,000 kilowatt (use 3/4). 

1 kilowatt = 1,000/746 horsepower (use 4/3). 

SPECIFIC GRAVITY OF COMMON SUBSTANCES 

Solids 
Aluminum (hard drawn) .... 2.7 

Brass (cast) 8.4 

Copper (cast) 8.9 

Gold (cast) 19.3 

Ice 918 

Iron (gray, cast) 7.1 

Iron (wrought) 7.8 

Nickel 8.6 

Platinum 21.4 

Silver (cast) 10.5 

Tin (cast) 7.3 Water (distilled).. . . 1.00 

Zinc (cast) 7.1 Water (sea) 1.03 

TENSILE STRENGTH ^ 

Cast iron 18,000 lb. /in. 2 

Wrought iron 50,000 lb./in.2 

Mild steel 70,000 lb./in.2 

COMPRESSIVE STRENGTH 

Cast iron 90,000 lb. /in. 2 

Wrought iron 38,000 lb. /in. 2 

Mild steel 80,000 lb./in.2 

SHEARING STRENGTH 

Cast iron 25,000 Ib./in.^ 

Wrought iron 40,000 lb./in.2 

Mild steel 50,000 lb./in.2 



Fluids 




Air 


00129 


Alcohol 


.80 


Chloroform 


1.5 


Ether 


.736 


Gasoline 


.66-. 6 


Kerosene 


.79 


Linseed oil (boiled) . 


.94 


Mercury 


13.6 


Neat's foot oil 


.91 


Turpentine 


.87 



APPENDIX 



217 



MODULUS OF ELASTICITY FOR TENSION AND COMPRESSION 

Cast iron 15,000,000 lb./in.2 

Wrought iron 25,000,000 lb./in.2 

Mild steel 30,000,000 lb./in.2 

MODULUS OF ELASTICITY FOR SHEAR 

Cast iron 6,000,000 Ib./in.^ 

Wrought iron 10,000,000 lb./in.2 

MUd steel 12,000,000 Ib./in.^ 

DECIMAL EQUIVALENT PARTS OF AN INCH 



3^4 0156 

M2 0313 

%4 0469 

He 0625 

%4 0781 

^2 0938 

K4 1094 

Ys 125 

1406 

1563 

1719 

He 1875 

i%4 2031 

M2 2188 

1^4 2344 

K 25 



764 
IK4 



IT 



64. 

^2' 
^4' 



2656 

2813 

2969 

%6 3125 

23^4 3281 

13^2 3438 

2^4 3594 

% 375 

2^4 3906 

1^2 4063 

2^4 4219 

Ke •• .4375 

2%4 4531 



15. 



32 



31 



64 

V2 



.4688 
.4844 
.5 



3a 



64 



5156 

5313 

^4 5469 

%6 5625 

3%4 5781 



IK2 

35. 



1^ 
3 9. 



32 

^4 



11 



'32 

u 

ie 



5938 

6094 

625 

6406 

6563 

6719 

6875 

4^4 7031 

2^2 7188 

4J^4 7344 

K..- 75 

4%4 7656 

2^2 7813 

53^4 7969 

^Ke 8125 

5%4 8281 

2^2 8438 

5%4 8594 

% 875 

s%4 8906 

2%2 9063 

59. 



^4 9219 

1^6 9375 

63^4 9531 

33^2 ". 9688 

6^^4 ■ 9844 

1 1 . 



218 



MECHANICS 
Trigonometric Functions 



Sin 



Cos 



Tan 



Sin 



Cos 



Tan 






0.000 


1.000 


0. 000 1 










1 


0.017 


0.999 


0.017 


46 


0.719 


0.695 


1.04 


2 


0.035 


0.999 


0. 035 ! 


47 


0.731 


0.682 


1.07 


3 


0.052 


0.999 


0.052 j 


48 


0.743 


0.669 


1.11 


4 


0.070 


0.998 


0. 070 


49 


0.755 


0.656 


1.15 


5 


0.087 


0.996 


0.087 


50 


0.766 


0.643 


1.19 


6 


0.105 


0.995 


0.105 


51 


0.777 


0.629 


1.23 


7 


0.122 


0.993 


0.123 


52 


0.788 


0.616 


1.28 


8 


0.139 


0.990 


0.141 


53 


0.799 


0.602 


1.33 


9 


0. 156 


0.988 


0. 158 


54 


0.809 


0.588 


1.38 


10 


0.174 


0.985 


0. 176 ; 


55 


0.819 


0.574 


1.43 


11 


0.191 


0.982 


0. 194 


56 


0.829 


0.559 


1.48 


12 


0.208 


0.978 


0.213 


57 


0.839 


0.545 


1.54 


13 


0.225 


0.974 


0.231 


58 


0.848 


0.530 


1.60 


14 


0.242 


0.970 


0. 249 


59 


0.857 


0.515 


1.66 


15 


0.259 


0.966 


0. 268 


60 


0.866 


0.500 


1.73 


16 


0.276 


0.961 


0.287 


61 


0.875 


0.485 


1.80 


17 


0.292 


0.956 


0.306 


62 


0.883 


0.469 


1.88 


18 


0.309 


0.951 


0. 325 


63 


0.891 


0.454 


1.96 


19 


0.326 


0.946 


0.344 


i 64 


0.898 


0.438 


2.05 


20 


0.342 


0.940 


0.364 


65 


0.906 


0.423 


2.14 


21 


0.358 


0.934 


0.384 


1 66 


0.914 


0.407 


2.25 


22 


0.375 


0.927 


0.404 


67 


0.921 


0.391 


2.36 


23 


0.391 


0.921 


0.424 


58 


0.927 


0.375 


2.48 


24 


0.407 


0.914 


0.445 


69 


0.934 


0.358 


2.61 


25 


0.423 


0.906 


0.466 


70 


0.940 


0.342 


2.75 


26 


0.438 


0.898 


0.488 


71 


0.946 


0.326 


2.90 


27 


0.454 


0.891 


0.510 


72 


0.951 


0.309 


3.08 


28 


0.469 


0.883 


0. 532 


1 73 


0.956 


0.292 


3.27 


29 


0.485 


0.875 


0. 554 1 


74 


0.961 


0.276 


3.49 


30 


0.500 


0.866 


0. 577 1 

1 


75 


0.966 


0.259 


3.73 


31 


0.515 


0.857 


0.601 


76 


0.970 


0.242 


4.01 


32 


0.530 


0.848 


0.625 


77 


0.974 


0.225 


4.33 


33 


0.545 


0.839 


0.649 


i 78 


0.978 


0.208 


4.70 


34 


0.559 


0.829 


0.675 


i 79 


0.982 


0.191 


5.14 


35 


0.574 


0.819 


0.700 


80 


0.985 


0.174 


5.67 


36 


0.588 


0.809 


0.727 


81 


0.988 


0.156 


6.31 


37 


0.602 


0.799 


0.754 


82 


0.990 


0.139 


7.12 


38 


0.616 


0.788 


0.781 


1 ^^ 


0.993 


0.122 


8.14 


39 


0.629 


0.777 


0.810 1 


84 


0.995 


0.105 


9.51 


40 


0.643 


0.766 


0. 839 


85 


0.996 


0.087 


11.43 


41 


0.656 


0.755 


0.869 


86 


0.998 


0.070 


14.30 


42 


0.669 


0.743 


0.900 


87 


0.999 


0.052 


19.08 


43 


0.682 


j 0.731 


0.933 


88 


0.999 


0.035 


28.64 


44 


0.695 


0.719 


0.966 


89 


0.999 


0.017 


57.28 


45 


0.707 


0.707 

1 


1.000 


90 


1.000 


0.000 


Infinity 



INDEX 



Absolute motion, 35 
Acceleration, 36 

of gravity, 25, 37, 203 
Acme thread, 147 
Adhesion, 5 
Aeroplane, 190 
Air, 180 

brake, 194 

chamber, 196 

pump, 193 
American system of rope trans- 
mission, 155 
Amplitude, 211 
Aneroid barometer, 184 
Annealing, 6 

Annular ball bearing, 108 
Anti-friction devices, 106 
Apex, 67 
Archimedes, 177 

law of, 177 
Areas, 215 
Arm, couple, 60 

moment, 33 
Ascensional force^ 191 
Atmosphere, 180 
Atmospheric pressure, 180 
Atom, 2 

Automobile, 135 
Axis of rotation, 32, 37 

B 

Babbitt metal, 106 
Ball bearing, 106, 107 
Balloon, 191 



Baltimore truss, 68 
Barograph, 185 
Barometer, mercurial, 183 

aneroid, 184 
Bearing, 106 

Bearings, lubrication of, 110 
Belts, 150 

Bevel gear, 155, 156, 157 
Bicycle, 130 
Block, 124 

Block and tackle, 117 
Block, chain, 148 
Bole, 38 
Boom, 73 
Bottom slack, 150 
Boyle, Robert, 188 
Boyle's law, 188 
Brace, 67 
Brake, prony, 93 

horsepower, 93 
Bridge, 67 

truss, 67 
British thermal unit, 100 
Brittleness, 6 
Buoyant force, 176 
Butt joint, 84 



Caliper, 12, 13 

micrometer, 14, 15 
vernier, 16 
Cam, 145 
follower, 141 
shaft, 141 
Capacity, metric table of, 10 
units of, 9 



219 



220 



INDEX 



Center of gravity, 25 

pressure, 170 
Centigram, 8 
Centiliter, 9 
Centimeter, 8 
Centrifugal force, 208 

pump, 197 
Centripetal force, 208 
C.g.s. sj'stem, 7 
Chain and sprocket, 147 
Chain block, 124 
Chemical change, 3 
Chord, 67 
Circle, area, 215 

circumference, 215 
Clutch, 144 

Coefficient of friction, 104 
Cohesion, 4 
Component, 44 
horizontal, 49 
vertical, 49 
Components, rectangular, 49 
Composition of forces, 44, 65 

velocities, 46 
Compound drive, 151 

truss, 66 
Compressibility, 4 
Compression, 32, 64, 66 

member, 66 
Compressive strength, table, 216 
Concurrent forces, 52 
Connecting rod, 106 
Conservation of energy, 100 
Continuous system of rope trans- 
mission, 155 
Cosecant, 20 
Cosine, 20, 218 
Cotangent, 20 
Countershaft, 136, 141 
Couple, 60 

arm, 60 
Coupling, 142 
Crane, 64 

hoisting^ 73 



Crankshaft, 140 
Cream separator, 208 
Crow bar, 114 
Cup-and-cone bearing, 107 
Curved-tooth gear, 157 
Curvilinear motion, 36 
Cutting tools, 122 
Cylinder, lateral area, 215 
volume, 215 

D 

Dam, 171 

Day, mean solar, 9 

Dead load, 66 

Decigram, 8 

Deciliter, 9 

Decimal equivalents, 217 

Decimeter, 8 

Deck truss, 66 

Dekameter, 9 

Density, 162 

table of, 215 

of water, 9 
Diagonal, 67 
Diagram, engine, 90 

indicator, 90 
Differential block, 124 
Double shear, 85 
Ductility, 6 
DjTie, 87 

E 

Efficiency, 113, 123, 126, 128, 131, 
134 

mechanical, 96 
Elastic fatigue, 83 

limit, 79 
Elasticity, 4, 77 

modulus of, 81, 217 
Electron, 2 
Energy, 2, 98 

conservation of, 100 

kinetic, 99 



INDEX 



221 



Energy, potential, 98 

transformation of, 100 
Engine diagram, 90 
English and metric equivalents, 
215 
system of measurement, 7 
rope transmission, 154 
Equilibrant, 46 
Equilibrium, 3, 27, 52, 57, 62 
Equivalents, metric and English, 
10, 215 
of heat, mechanical, 100 
Erg, 87 



Factor of safety, 82 
Fafnir ball bearing, 108 
Failure of rivets, 85 
Falling bodies, 202 
Fatigue, elastic, 83 
Female thread, 147 
Fink truss, 67 
First class lever, 114 
Fishing rod, 115 
Fixed coupling, 142 

pulley, 117 
Flexible coupling, 143 

shaft, 142 
Foot-pound, 88 
Force, 30, 35 

centripetal, 208 

centrifugal, 208 

graphical representation of, 31 

measure of, 31 

moment of, 32 

of gravity, 24 

pump, 195 
Forces, composition of, 44, 65 

concurrent, 52 

non-concurrent, 62 

parallelogram of, 45 

triangle of, 53 
F.p.s. system, 7 



Friction, 103 

clutch, 144 

coefficient of, 104 

laws of, 105 

sliding, 104 

wheels, 155 
Fulcrum, 32, 57, 114, 115 
Functions, trigonometric, 20, 218 
Fundamental units, 7 



G 



^^g," 215 

Galileo, Galilei, 182 
Gas, 161, 180 
Gauge, pressure, 198 

vacuum, 198 
Gear, 156 

transmission, 135 

reverse, 136 

pump, 197 
Geometrical formulas, 215 
Governor, 209 
Gram, 8 

Graph, 126, 129, 132, 135 
Graphical representation of forces, 

31 
Gravitation, 23 

law, 23 
Gravity, 24 

acceleration of, 25, 37, 203 
Ground reaction, 63 | 

h; 

Hardness, 5 

Heat, mechanical equivalent, 100 

Hectometer, 9 

Helical gear, 156, 157, 159 

Herringbone gear, 159 

High-duty bearing, 106 

Highway bridge, 67 

Hoisting crane, 73 

Hollow shaft, 140 



222 



INDEX 



Hooke, Robert, 77 
Hooke's law, 77, 82 
Horizontal, 24 

component, 49 
Horsepower, 92, 152 

indicated, 95 
Howe truss, 67 
Hyatt bearing, 109 
Hydraulic press, 166 
Hydrometer, 178 

I 

Idler, 150, 151 

reverse, 137 
Inclined plane, 112, 120 
Indestructibility, 4 
Indicated horsepower, 95 
Indicator, 96 

diagram, 90 
Inertia, 4 

law of, 39 
Input, 112 
Inside caliper, 13 
Instantaneous velocity, 37 

J 

Jack screw, 121, 132 

Joint, 84 
Joule, 88 

James Fresco tt, 101 

K 

Kelvin, Lord, 2 
Kilogram, 8 
Kilogram-meter, 88 
Kilowatt, 92, 93 
Kinetic energy, 99 

theory, 3 
Kinetics, 1 



Ladder, 62 



Lap joint, 84 

Lathe, lead screw, 147 

Law, Archimedes', 177 

Boyle's, 188 

gravitation, 23 

Hooke's, 82 

inertia, 39 

machines, 113 

moments, 57 

parallelogram, 44 

Pascal's, 165 
Laws, friction, 105 

liquid pressure, 164 

motion, 38 

pendulum, 213 
Lead, 121, 147 
Length, measures of, 9 

standard, English, 7 
metric, 8 
Lever, 112, 114, 115 
Lift, 191 

pump, 195 
Line shaft, 140 
Linear table, metric, 9 
Link, 146 

liquid, 161 
Liquid pressure, computation, 170 

laws, 164 
Liter, 9 
Live load, 66 
Load, dead, 66 
Low-duty bearing, 106 
Lubrication, 110 



M 



Machines, 112 

laws of, 113 
Magdeburg hemispheres, 186 
Main shaft, 140 
Male thread, 146 
Malleability, 6 

Manila transmission rope, 154 
Manometer, 199 



INDEX 



223 



Mass, 5 

standard of, English, 7 
metric, 8 
Matter, 2 
Mean solar day, 9 
Measurement of a force, 31 
Measures, metric, tables of, 9, 10 
Measuring instruments, 11 
Mechanical advantage, 113 

efficiency, 96 

equivalent of heat, 100 
Mechanics, 1 

of liquids, 163 
Medium-duty bearing, 106 
Member, compression, 66 

tension, 66 
Mercurial barometer, 183 
Meter, 8 

Metric and English equivalents, 
215 
tables, 9, 10 

system, 7 
Micrometer caliper, 14, 15 
Milligram, 8 
Milliliter, 9 
Millimeter, 8 
Mitre gear, 157 

Modulus, of elasticity, Young's, 
81, 217 

of rigidity, 84 
Molecule, 2 
Moment of a force, 32 

arm, 33 
Moments, law of, 57 
Momentum, 38 
Motion, 35 

laws of, Newton's, 38 
Multiple system of rope trans- 
mission, 154 
Myriameter, 9 



N 



Neutral equilibrium, 27 



Newton, Sir Isaac, 23 
Newton's laws of motion, 38 
Non-concurrent forces, 62 



Output, 113 

Outside caliper, 12, 13 

Overshot wheel, 174 



Panel, 67 
Parallel forces, 57 
Parallelogram law, 44 

of forces, 45 
Pascal's law, 165 
Pel ton wheel, 175 
Pendulum, 210 

laws of, 213 
Penstock, 175 
Period, 211 
Physical change, 3 
Physics, 1 
Pinion, 157 
Pitch, 121, 147 
Planetary system, 130 
Plumb line, 24 
Porosity, 3 
Positive clutch, 144 
Potential energy, 98 
Power, 92 
Pratt truss, 67 
Pressure, 162 

atmospheric, 180 

gauge, 198 

liquid, 163 
Projectiles, 204, 205 
Prony brake, 93 
Properties of matter, 3 
Protractor, 12 
Pulley, 112, 117 

whip-on-whip, 139 
Pulleys and ropes, 153 



224 



INDEX 



Pump, air, 193 
force, 195 
lift, 195 

reciprocating, 195 
rotary, 197 

Q 

Quadrangular truss, 68 

R 

Radial bearing, 106 
Radian, 41 
Railroad bridge, 67 
Range, 206 
Reaction, 31 

ground, 63 
Reciprocating pump, 195 
Rectangular components, 49 
Rectilinear motion, 36 
Representation of forces, graphic, 

31 
Resolution of forces, 48 

velocities, 48 
Resultant, 44, 46 

parallel forces, 59 
Retaining walls, 171 
Reverse gear, 136 

idler, 137 
Rigidity, modulus of, 84 
Rivets, failure of, 85 
Roller bearing, 106, 109 

chain, 148 
Roof truss, 67, 68 
Rotary motion, 36 

pump, 197 
Rotation, 36 

axis of, 32 



S 



Safety factor, 82 
Screw, 112, 121 



Screw, threads, 146 
Screw-geared block, 124, 127 
Secant, 20 

Second class lever, 115 
Secondary shaft, 142 
Shaft, 140 

counter, 136 

spline, 136 
Shear, 83 

double, 85 

single, 85 
Shear legs, 74 
Shearing strain, 84 

strength, table, 216 

stress, 84 
Silent chain, 149 
Simple harmonic motion, 211 

machines, 112 
Sine, 20, 218 
Siphon, 193 
Slide rule, 14 
Sliding friction, 104 

gear, 135 
Society Automotive Engineers, 95 
Solid, 161 

Specific gravity-, 162, 178, 216 
Speed, 37 

indicator, 13, 14 
Sphere, area, 215 

volume, 215 
Spiral gear, 156, 169 
Spline shaft, 136 
Sprocket, 147 
Spur gear, 155, 156, 157 
Spur-geared block, 124, 129, 130 
Square thread, 147 
Stability, 28 
Stable equilibrium, 27 
Standard of length, English, 7 
metric, 8 
mass, English, 7 

metric, 8 
time, 7 

conditions, 186 



INDEX 



225 



Statics, 1 

Stay, 67 

Steam engine diagram, 90 

Steel rule, 11 

Stick and tie, 70 

Strain, 81 

shearing, 84 
Stress, 80 

shearing, 84 
Strut, 67 

Surface, metric table, 10 
System, c.g.s., 7 

f.p.s., 7 

metric, 7 



Trigonometry, 20 
Trusses, 66 
Turbine wheel, 175 
Turnbuckle, 138 

U 

Undershot wheel, 175 
Units, 7 

of power, 92, 93 

of work, 87 
Universal coupling, 143 

gravitation, 23 
Unstable equilibrium, 27 



Tables, 215 

metric, 9, 10 
Tail race, 176 
Tandem drive, 151 
Tangent, 20, 218 
Tempering, 6 
Tenacity, 5 

Tensile strength, 5, 82, 216 
Tension, 32, 64, 66 
Theory, kinetic, 3 
Third class lever, 115 
Thomson, Sir William, 2 
Thrust bearing, 106 
Tie, 67 

Time, standard of, 7 
Timken bearing, 109, 110 
Tools, cutting, 122 
Toothed gear, 156 
Top slack, 150 
Torricelli, 182 
Trajectory, 206 
Transformation of energy, 100 
Translator y motion, 36 
Transmission, 140 

gear, 135 
Triangle of forces, 53 
Trigonometric functions, 20, 218 



Vacuum, 182, 202 

gauge, 198 
Valves, 141 
Vector, 41 

Velocities, composition of, 46 
Velocity, 37 

instantaneous, 37 

ratio, 113, 125, 128, 131, 134 
Vernier caliper, 16 
Vertical, 24, 67 

component, 49 
Vise, 138 
Volume, metric table, 10 

of a cylinder, 215 

of a sphere, 215 
Von Guericke, Otto, 186 
^^V" thread, 147 

W 

Wall crane, 64 
Water, density of, 9 

wheel, 174 
Watt, 93 

James, 92 
Web member, 67 
Wedge, 112, 122 



226 



INDEK 



Weight, 24 

metric table of, 10 
Wheel and axle, 112, 119 
Wheelbarrow, 115 
*'Whip-on-whip" pulley, 139 
Wire transmission rope, 154 
Work, 87 

diagram, 89 

units, 87 



Worm, 127 

gear, 156, 158 
wheel, 127 



Yard, 8 

Yield point, 80 
Young's modulus of elasticity, 81. 
217 



t 



